Newton’s Laws, Part 5

Please read the prior posts in this series if you have not been following along. I ended last post with two equations, each relating to a different phase of our model rockets flight: powered and coasting. Let’s look at the powered phase (phase 1).

For this phase, Newton’s second law (F = ma) reduced down to:

5.08 = 0.4a

From this equation, a first class rocket scientist can use calculus to find the velocity and the distance from the launch pad at any time in seconds after launch. These equations are:

v(t) = 12.7t, x(t) = 6.35t²

where v(t) is the velocity t seconds after launch and x(t) is the distance from the launch pad t seconds after launch. Now I’ve introduced what is called functional notation. Instead of saying the velocity or distance after 3 seconds, I can just say v(3) or x(3). Maths is full of shorthand notations.

These two equations assumes that we start the clock at 0 seconds and that velocity, acceleration, and distance are all 0 at 0 seconds. Now remember, these equations are only valid for the first 3.3 seconds of flight (see previous post) because the engine stops burning at 3.3 seconds.

So how fast is our rocket going at 3.3 seconds? Well, just replace t with 3.3 in the velocity equation and calculate it:

v(3.3) = 12.7×3.3 = 41.91 m/s

To give you a perspective of how fast this is, this is equivalent to almost 151 km/h. How high is the rocket at engine burnout? Let’s replace t with 3.3 in the distance equation and calculate it:

x(3.3) = 6.35×(3.3)² = 69.15 m

Is this the highest the rocket goes, a measly 69 meters? Well remember, at burnout, the rocket is going up very fast. It will take gravity a while to turn that around. Enter the phase 2 equations.

From my last post, Newton’s second law for phase 2 is:

a = -9.8

Again, using calculus, our friend, the first class rocket scientist generates the two equations (applicable only to phase 2):

v(t) = -9.8t + 74.25, x(t) = -4.9t² + 74.25t -122.514

These equations are a bit more complex because they have to take into account that at 3.3 seconds, the velocity is 41.91 m/s and the rocket is 69.15 m high.

So how high does our rocket go? At the peak of its travels, the velocity goes from positive (going up) to negative (going down). That is, it passes through 0. So in order to find the highest that our rocket goes, we need to find when the velocity equals 0. So we use our velocity equation and set it equal to o, then solve for the time that makes that happen:

v(t) = -9.8t + 74.25 = 0

-9.8t = -74.25

t = -74.25/-9.8 = 7.58 seconds

So now we use the distance equation and replace t with 7.58:

x(7.58) = -4.9×(7.58)² + 74.25×7.58 -122.514 = 158.76 m

So now remember that we are not using a parachute. So the next two questions to ask is when does it hit the ground and how fast is it going when it does.

When the rocket hits the ground, its distance is 0. So now we use the distance equation, set it equal to 0 and find the value of t to make that happen:

x(t) = -4.9t² + 74.25t -122.514 = 0

Now you can solve this using the quadratic formula which I have covered in a previous post. Using this formula, you get two answers: 1.884s and 13.269s. The first answer is not greater than 3.3. These phase 2 equations are only valid for t greater than 3.3 seconds. So we can reject that answer and choose 13.269 seconds. So the total flight time is a bit over 13 seconds.

Now how fast does it hit the ground? Put the time 13.269 into the velocity equation to get:

v(13.269) = -9.8×13.269 + 74.25 = -55.79 m/s

The velocity is negative because it is going down. So the rocket is going its fastest when it hits the ground, not when the engine burns out. 55.79 m/s is equivalent to 200.84 km/h. What are the odds that we can launch this rocket again?

In my next post, let’s do the same problem but use a bigger rocket (since our model rocket is now one with the earth).

Newton’s Laws, Part 4

Well congratulations! If you get through (and understand) this and the last three posts on Newton’s laws, consider yourself a rocket scientist third class.

In my high school days, my friend Byron and I formed a two-member rocket club. We would pool our money and get and build model rockets. We also got solid propellant cartridges that were inserted into the rocket. We would put the rocket on a launch pad (a wooden block) and ignite the propellant, either with a fuse or electrically with a small coil of wire that was heated with an electric current. The rocket would ascend very fast, then a small charge at the end of the propellant cartridge would push out the nose cone and the folded parachute inside. Then the fun part was to try and retrieve the rocket, especially on windy days!

So today, I am going the find the equation of motion and the velocity equation of these rockets using Newton’s second law, F = ma.

We are going to make several assumptions. I will relax some of these later, but seeing that you are only a third class rocket scientist, let’s start out making things as simple as possible.

Assumptions:

  1. The force exerted by the cartridge is constant from ignition to the time it cuts off.
  2. The mass of the rocket is constant throughout its flight.
  3. The rocket is going straight up and down. (We will not be using a parachute.)
  4. The flight time is short so we will not take into account the rotation of the earth.
  5. The force due to gravity is constant throughout the flight.
  6. Air drag is ignored.

A first class rocket scientist would not be making these assumptions and would include the effects of these into the equations. Assumption 1 is not valid for real rockets and the varying force would have to be accounted for.

Assumption 2 would make the equations very inaccurate in reality. Depending on the type of propellant, the rocket fuel can be from 83% to 96% of the total mass of the rocket. As the fuel is burned, the mass of the rocket is getting less and less, and as you can see from F = ma, this means that the acceleration must increase if m×a is to continue equalling the constant force.

Assumption 3 is made so that the velocity, acceleration, and position are measured along the same line. For big rockets, measurements of these three things are measured with something called vectors which provide a direction as well as a value.

Assumption 4 is made because the earth does rotate and this rotation adds to the speed, not in the vertical direction but in a sideways direction.

Assumption 5 is OK for our model rocket because it will not go that high, but the force of gravity does lessen with altitude and this must be taken into account for larger rockets.

Assumption 6 allows us to keep the equations simple (as do all of them).

Now let’s look at an actual model rocket. These numbers are realistic numbers that are possible for model rockets.

Suppose we launch a model rocket that weighs 400 g. The engine provides 9 N of force for 3.3 s. Let’s answer the following questions:

  1. How high does the rocket go?
  2. What is the maximum speed of the rocket?
  3. What is the total flight time?

Before I set this up, let’s agree on what’s positive and what’s negative. Let’s agree that up is positive and down is negative. So the force due to the rocket engine is positive because it is pushing the rocket up and the force due to gravity is negative because it is pulling the rocket down.

Now there are two phases of the rocket’s flight: the first phase is when the engine is burning and the second phase begins when the engine stops. Let’s look at F = ma during the first phase.

Phase 1 – engine is burning: There are two forces acting on the rocket, the one due to the engine and the other due to gravity. In part 2 of this series of posts, I explained how to find the force due to gravity. The gravity force points down so it is negative. The other force, of course, is due to the rocket engine and it is positive. So putting this into F = ma:

9 N – (0.4 kg)(9.8 m/s²) = 0.4 kg × a

I use 0.4 kg instead of 400 g because we are using SI units (explained in last post). So simplifying this equation and removing the units gives:

5.08 = 0.4a . This is the starting equation for the first phase.

Phase 2 – engine stops: Now let’s look at the second phase. After the engine cuts off, the only force acting on the rocket is gravity. So the only acceleration the rocket is experiencing is only due to gravity. So F = ma becomes:

– (0.4 kg)(9.8 m/s²) = 0.4 kg × a

But you can see from this that a = -9.8 m/s² to make this a true equation. This simple equation is the starting equation for the second phase.

This post is now longer than I thought it would be, so I will continue with this in my next post.

Newton’s Laws, Part 3

What I would like to eventually get to, is to develop the equation of motion of a rocket. An equation of motion is just an equation that calculates an object’s position given a time. I did this without a lot of detail, in my Springy Thingy posts back in February. For this set of posts, I would like to add a bit more development.

Let’s go back to our basic equation that describes motion: F = ma. Let’s look at the a (acceleration) part. Acceleration is a rate of change of velocity. If a car goes from 0 to 100 km/s in 10 seconds, its acceleration is the difference in velocity divided by the time interval:

\[
{a}\hspace{0.33em}{=}\hspace{0.33em}\frac{{100}\hspace{0.33em}{\mathrm{k}}{\mathrm{m}}{/}{\mathrm{s}}}{{10}\hspace{0.33em}{\mathrm{s}}}\hspace{0.33em}{=}\hspace{0.33em}{10}\hspace{0.33em}\frac{\mathrm{k}\mathrm{m}/\mathrm{s}}{\mathrm{s}}\hspace{0.33em}{=}\hspace{0.33em}{10}\hspace{0.33em}{\mathrm{k}\mathrm{m}/\mathrm{s}}^{2}
\]

This means the car increases its velocity 10 km/s each second. So acceleration is the rate that velocity changes per unit of time. However, velocity is also a rate of change measurement. Velocity is the rate of change of position (or distance) per unit of time. An equation of motion is finding the position of an object given a time.

So acceleration is the rate of change of velocity which is the rate of change of position, and it’s position that we want. How do we get there? This is where calculus comes in.

Calculus essentially deals with rate of change equations. It can find the rate of change of position, that is velocity, given a position equation. It can also go backwards and find a position equation, given a velocity equation. In our case, it can take the rate of change of velocity (acceleration) and find the velocity equation and then take the velocity equation and find the position equation, that is the equation of motion.

This ability to take a simple equation like F = ma, and from it, describe the motion of objects is one of the many reasons I love maths.

With this as a background, next time, let’s launch a rocket, then cut off it’s motors and answer questions like:

How high is the rocket when the motors cut off?

How fast is the rocket going when the rockets cut off?

How high does the rocket go?

When will the rocket hit the ground (or will it hit the ground)?

Newton’s Laws, Part 2

So I left off last time with the equation form of Newton’s second law: F = ma. It is important to use consistent units for the three quantities, the force F, the mass m, and the acceleration a.

Now in the USA, they use English units where the unit for acceleration is ft/sec² (feet per second squared), the unit of mass is something called a slug, and the unit for force is lbf (pound-force). We will not be using these units.

In the civilised world (I’m not biased), we use SI units. SI comes from the French Système international  which means the international system of units. For our equation, these units are kg (kilogram) for mass, m/sec² (meters/second squared) for acceleration, and the combination of these units on the right side of F = ma gives kg×m/sec² as the unit of force. This unit is given a special name in SI units, the newton, N, in honour of guess who. So 1 N is the force required to accelerate 1 kg of mass, 1 m/sec². 1 N is about the force an average size apple exerts on your hand.

Now I said I would also explain the difference between mass and weight. Weight is a force exerted by an object due to gravity. An objects weight changes when measured on different planets or moons. Its mass however, is an intrinsic property and remains unchanged regardless of where the object is. Mass is the amount of stuff that makes up the object.

Now the confusion between these two things arises because we commonly use weight, say in kilograms, to mean force. We feel the weight of a 1 kg object in our hands. But this unit is really a kilogram-force (kgf). A kilogram in SI units is a unit of mass, not weight. But fortunately, an object on earth that exerts a force of 1 kgf due to gravity, is defined as having a mass of 1 kg, so it is easy to interchange these units on earth. But elsewhere, the object will exert a different force (kgf) due to gravity but it will still be an object of 1 kg mass.

The kilogram-force unit is not an SI unit – we just naturally use it in everyday conversation. As seen before, the newton (N) is the SI unit of force. So what is the weight of a 1 kg mass in SI units? On earth, when you drop something from a height, its velocity starts as 0 m/s but its speed increases as time goes on. This is why you can jump from a half a meter height without injuring yourself but will do considerable damage if you jump from a 50 meter height. Increasing speed means acceleration. So there is an acceleration due to gravity on earth as well as other planets. On earth, the acceleration due to gravity is 9.8 m/sec². When an object is dropped on earth, its speed increases by 9.8 m/sec every second.

So if we want to calculate the weight of a 1 kg object, we use Newtons second law F = ma where we use the acceleration due to gravity for a:

F = 1 kg × 9.8 m/sec² = 9.8 N

If you want to know your weight in newtons, just take your mass in kg (which equals your weight in kgf), and multiply by 9.8. Your weight in newtons is almost 10 times your weight in kgf, which explains why people prefer to use kgf.

Newton’s Laws, Part 1

As my interest in maths grew out of seeing it applied, I thought I should start writing posts on its applications. Physics applications are almost always maths related and you can’t start a conversation about physics without starting with Newton’s Laws.

Sir Isaac Newton is famous for his work in physics and maths. I find it amazing that his accomplishments occurred in the 1600’s. One of his most famous works, some would say his most famous, was his Principia Mathematica Philosophiae Naturalis (The Mathematical Principles of Natural Philosophy). In his day, the term natural philosophy meant science. In this publication, Newton set out his three laws of motion:

  1. Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.
  2. Force equals mass times acceleration.
  3. For every action there is an equal and opposite reaction.

The first law is sometimes called the law of inertia. You experience inertia every day when you try to push an object or stop an object from moving. You have to apply a force to start an object moving or stop one from moving. The second law explains why a greater force is needed to stop a moving car than a baby stroller. The third law explains why rockets work, what happens when you release a ballon before it’s tied, and why a gun or rifle has a recoil.

This series of posts will be mainly about Newton’s second law. This law, in equation form, is

F = ma , where F is force, m is mass, and a is acceleration.

Before we work with this equation using numbers, let’s see what this equation means.

If you try to stop a rolling car, you are trying to decrease its velocity from a certain value to zero. In other words, you are trying to decelerate it. Deceleration is negative acceleration, and according to Newtons second law, because the mass of the car is rather large, a large force is required to stop it from rolling. A baby stroller going the same speed requires less force because its mass, m, is much smaller than a car’s mass.

Even though this equation is very simple, there are entire books dedicated to this equation. The rest of my posts could easily be about this equation alone, but I will try to keep it down to just a few.

In my next post, I will talk about the units that we will use for the three things that make up this equation. I will also talk about the difference between mass and weight.

Logarithms, Part 5

I never thought these posts would get to 5.

Now I said I would do a population problem but I have decided to go with a radioactive decay problem instead. I will use the example of carbon dating as this is based on radioactive decay. But first, let’s look at the general equation for exponential decay:

\[
{A}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{kt}}
\]

This formula gives the amount of something that is decreasing exponentially. A is the amount left after t seconds, hours, days, years or whatever depending on the value of the rate of decrease factor which is k. A0 is the amount of something we started out with, the amount present at t = 0. This formula makes sense when you look at the ekt part of the equation.

Now I have talked about e before. It is an irrational number, like 𝜋, and is approximately equal to 2.7183. k is a rate of decrease factor that depends on the material we are working with and the units of time t. The –kt part, the exponent of e, is a negative number since k and t are positive. I have explained negative exponents before but ekt equals 1/ekt . Now what happens to ekt as t gets large? Any number greater than 1 (which e is) raised to a larger and larger power, gets very big. And when you divide a big number into 1, you get a very small number. So A0 is being multiplied by a number that gets smaller and smaller as time goes on. That is why A, the amount of material, is exponentially decreasing.

With that as a background, let’s talk about carbon dating. Any living thing has carbon in it. Indeed, all life on earth is carbon-based which means that the the molecules essential for life are composed of lots of carbon. Now carbon comes in different “flavors”. These flavors are called isotopes and carbon has two main isotopes: carbon 12 the most abundant and non-radioactive, and carbon 14 which is radioactive. Fortunately, the amount of carbon 14 is very small – about 1 atom to every 1012 atoms of carbon 12. However, in living things, this ratio is pretty much constant since carbon 14 is continually made in our atmosphere. But once something dies, the carbon 14 is not replenished and the amount present at the time of death starts decreasing.

So carbon dating is a process of determining the amount of carbon 14 left in a once living object then calculating the time it would take to have that much carbon 14 left.

So let’s go back to our equation for exponential decay. In order to use this equation for carbon dating, we need to know what k is for carbon 14. Now we know that the half-life of carbon 14 is 5700 years which means that given any amount of carbon 14, only half that amount will be left in 5700 years due to radioactive decay. So let’s use this fact to calculate k.

Taking this information and putting it into our equation results in

\[
{0}{.}{5}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{k}\times{5700}}
\]

So the left side shows that there is half (0.5) the initial amount and the right side shows that this occurs in 5700 years. So now, I will take the loge (abbreviated as ln) of both sides. Note that since e is the base on the right side, taking the log to that base just results in the exponent –kt. Also note that A0 appears on both sides of the equation, so we can divide both sides of the equation by A0 which makes A0 disappear:

\[
\begin{array}{l}
{{0}{.}{5}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{k}\times{5700}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}{=}\hspace{0.33em}{e}^{{-}{k}\times{5700}}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\ln{(}{0}{.}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}\ln{(}{e}^{{-}{k}\times{5700}}{)}\hspace{0.33em}{=}\hspace{0.33em}{-}{k}\times{5700}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{0}{.}{6931}\hspace{0.33em}{=}\hspace{0.33em}{-}{k}\times{5700}}\\
{\Longrightarrow{k}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{0}{.}{6931}}{5700}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{0001216}}
\end{array}
\]

So now that we know what k is, we can use the following equation to do our carbon dating:

\[
{A}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{0}{.}{0001216}{t}}
\]

So let’s say a fossil has 35% (0.35) of its original carbon 14 when it died. How old is the fossil?

\[
\begin{array}{l}
{{0}{.}{35}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{0}{.}{0001216}{t}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{35}\hspace{0.33em}{=}\hspace{0.33em}{e}^{{-}{0}{.}{0001216}{t}}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\ln{(}{0}{.}{35}{)}\hspace{0.33em}{=}\hspace{0.33em}\ln{(}{e}^{{-}{0}{.}{0001216}{t}}{)}\hspace{0.33em}{=}\hspace{0.33em}{-}{0}{.}{0001216}{t}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{1}{.}{0498}\hspace{0.33em}{=}\hspace{0.33em}{-}{0}{.}{0001216}{t}}\\
{\Longrightarrow{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{1}{.}{0498}}{{-}{0}{.}{0001216}}\hspace{0.33em}{=}\hspace{0.33em}{8633}\hspace{0.33em}{\mathrm{years}}}
\end{array}
\]

We have a lot of birthdays to catch up on!

Logarithms, Part 4

Let’s do another example using logarithms. As seen in my last post, logarithms are useful when the unknown variable in an equation is in the exponent of some number. But the exponent can be more than just the unknown – it can be an expression with an unknown. Consider the following problem:

\[
{10}^{{3}{x}{+}{7}}\hspace{0.33em}{=}\hspace{0.33em}{125}
\]

So the first step, as seen last time, is to take the log of both sides of the equation. We then can use the property of logs that was introduced: \[ {\log}_{a}{b}^{x}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}{\log}_{a}{b} \]

So let’s again use the base 10 log, the log x key on your calculator:

\[
\begin{array}{l}
{{10}^{{3}{x}{+}{7}}\hspace{0.33em}{=}\hspace{0.33em}{125}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\log{(}{10}^{{3}{x}{+}{7}}{)}\hspace{0.33em}{=}\hspace{0.33em}\log{(}{125}{)}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{(}{3}{x}\hspace{0.33em}{+}\hspace{0.33em}{7}{)}\log{10}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{0969}}
\end{array}
\]

Now let’s stop here. The log of 125 is done on your calculator. What about the log of 10? Well that can be done on your calculator as well, but if you’ve been paying attention, you can see that the answer is 1. 1 is the exponent of 10 to make 10? = 10. So now we have a standard (non-exponential) equation:

\[
{3}{x}\hspace{0.33em}{+}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{0969}
\]

We have solved equations like this before, so without going into the detail, the solution to this is x = -1.6344. You can put this value of x in the left side of the original equation and find that it does solve it.

In my next post, I will present another property of logs and use it to solve a population problem.

Logarithms, Part 3

Finally have a little time for a post.

So we know how to solve x2 = 10 by taking the square root of both side of the equation to get x = ±3.162… Note that taking the square root of x2 undoes or reverses the squaring of x.

But what do you do if x is in the exponent and not the base?

\[
{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{10}
\]

You can’t take the xth root since you don’t know what x is. So what to do? From my last post, you saw that log2 10 means “what is the number that I can use as the exponent of 2 so that the answer is 10”. So in the above equation, if I take the log2 of both sides, I get

\[
{\log}_{2}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}
\]

The left side of this equation is doing two inverse operations on the number 2 – raising 2 to a power then taking its log. In other words, the left side can be seen as saying “what is the number that I can use as the exponent of 2 so that the answer is 2x ?”. Well the answer to that question is x. So the left side is just x and the right side is just a calculation:

\[
{\log}_{2}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}\hspace{0.33em}{=}\hspace{0.33em}{3}{.}{32192809488}
\]

Well that’s just dandy! Trouble is, without the internet, how do you find log2 10? I have not seen a calculator with a log2 x button. As mentioned in my last post, calculators usually have buttons to take logs relative to bases 10 and e. Well fortunately, there are lots of properties of logs that can help. The one we can use here is

\[
{\log}_{a}{b}^{x}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}{\log}_{a}{b}
\]

This means that I can take the log with respect to any base, and the x can be removed as the exponent. So for our problem, let’s take the log10 (the log x key on your calculator) of both sides and see what happens:

\[
{\log}_{10}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{10}{10}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}{\log}_{10}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}
\]

Let’s stop here for a moment before I complete the solution. Why is the right side equal to 1? Log10 10 is saying “what power of 10 equals 10?”. The answer is 1 because 101 = 10. On the left side, I used to log property above to bring the x in front of the log. Now log10 2 is just a number. You can use the log x key on your calculator to find that log10 2 = 0.3010 to four decimal places. So now the equation becomes

\[
{x}{\log}_{10}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{301}{x}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{0.301}\hspace{0.33em}{=}\hspace{0.33em}{3}{.}{3219}
\]

So 23.3219 = 10. In my next post on logs, I’ll do more equation solving using logs.

Logarithms, Part 2

So what is a logarithm? Let’s first see the notation, then I will explain. When taking the log (short for logarithm which I will use from now on) of a number, you need to know what base is being used. The notation for the log of x is loga x. The a is the base and is usually a specified number. so examples using this notation are log2 10, log10 25, log18 145, loge 7.34. Let’s look at these.

log2 10 is asking the question “What number can I use as the exponent of 2 so that the answer is 10?”. It turns out that 23.321928094887 = 10 so log2 10 = 3.321928094887.

log10 25 is asking the question “What number can I use as the exponent of 10 so that the answer is 25?”. Well, 101.39794 =25 so log10 25 = 1.39794.

Are you getting the picture? What about log18 145? This is asking the question “What number can I use as the exponent of 18 so that the answer is 145?”. 181.72183 = 145 so log18 145 = 1.72183.

Now let’s look at loge 7.34. This shows that the base or the number we are taking the log of does not have to be an integer. The number e, which I have talked about before, is an irrational number, but it still can be used as a base. In fact, it is probably the most used base. Since e1.99334 = 7.34, then it follows that loge 7.34 = 1.99334.

By the way, on most calculators, the log or log x key assumes that the base is 10. On most calculators as well, ln x means loge x. “ln” means “natural log”.

Now loga x and ax are inverses of each other. This means that one undoes the other. So if on your calculator, you find ln 7, then take that number and hit the ex key, you get the original 7 back. This works in reverse as well: Find e7 on your calculator, then hit the ln x key. You will again get the 7 back.

In notation-speak, this inverseness is shown as

\[
\begin{array}{l}
{{a}^{{\log}_{a}x}\hspace{0.33em}{=}\hspace{0.33em}{x}}\\
{{\log}_{a}{a}^{x}{=}\hspace{0.33em}{x}}
\end{array}
\]

In my next post, I will show how logarithms can be used to solve equations.