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### Course: Precalculus (Eureka Math/EngageNY)>Unit 3

Lesson 15: Topic C: Lessons 20-21: Inverse relationship of exponentials and logarithms

# Relationship between exponentials & logarithms: graphs

Given a few points on the graph of an exponential function, Sal plots the corresponding points on the graph of the corresponding logarithmic function. Created by Sal Khan.

## Want to join the conversation?

• Hi,

This might be a silly question but how is y = log_b (X) the inverse of y=b^x? Isn't the inverse of y=b^x this log_b (y) = X?

Thanks
• I guess if you look at the function as y = f(x) = b^x. The inverse of the function would be x = f(y) = b^y. So, the logarithmic form of the inverse function is log_b(x) = y.
Ey that's just my take, might be wrong too lol.
• How did he know to use points 1,4, and 16 in the second chart? I thought that, given the question, he would have taken the x values from the existing points and plugged them into the 2nd function.
• For anyone who watches this video and has a similar question, think about it this way:
y = b^x can be rewritten as log_b(y) = x.

So for b = 4 (you can deduce that from the graph) and y = 1, 4, 16 (also deduced from the graph) you have

log_4(1) = 0
log_4(4) = 1
log_4(16) = 2

Which represent points (1, 0), (4, 1) and (16,2) the inverse of the points on the given graph!
• at around , why is sal taking the y values of the first table and using them as the x values of the second table?
• I had a little trouble on this too, but here's what I figured: The purpose of an exponential function is to find what value a number equals when it is raised to the power of x. However, the *purpose of a logarithm is to find the exponent (x value)* that gives you a certain amount when you raise a base by the unknown x value. Though it might sound complicated, if you try to define the purpose yourself, you'll end up with the same answer. For this reason, they are like the inverse functions of each other, just like multiplication and division.

In other words, the logarithm tries to lead you to the exponent needed to reach the value, while the exponential graph tries to lead you to the value given by the exponent's use.

Therefore, they are inverse operations as they undo each other. Furthermore, they reflect over the y=x line, and their coordinates are switched. If you check at , you'll see Sal agrees. If you check the background at this time, you'll also see how when the b= 4, everything fits together perfectly.

Hope this helps.
• Sorry about this but I have two questions. The first might be silly, but here goes: How does one solve for x when it is an exponent such as 2^x=1/64. I've been having to guess this whole time and try for answers on my calculator, because solving for x has not been working. I think I might be doing it wrong...
Second is, I didn't understand this video too well. I don't understand what the graph has
to do with any of this, even though I've been okay with everything pertaining to logarithms so far. Any answer is much appreciated. Thank you!
• When the variable is in the exponent, you need to use logarithms of whatever the base of the exponent is.
For 2^x = 1 / 64, the base is 2. Therefore, we'll be taking log base 2 of each side of the equation.
But before doing that, it's usually easiest to express both sides of the equation using the same base.
So, 2^x = 1 / 64 = 1 / 2^6 = 2^(-6)
log_2 ( 2^x ) = log_2 ( 2^-6 )
x = - 6
Hope this helps.
(I'll let someone else pick up on your question about the graph.)
• why did he take the y values and plug them into the log function? How did he know not to just use the same x values for the log function as for the exponential function?
• This is because logarithmic functions and exponential functions are inverse functions of each other. The inverse of a function is the function that takes the range (y) of that function to the domain (x) of the function. Therefore, the domain (x) of the inverse function will be the range (y) of the original function and the range (y) of the inverse function will be the domain (x) of the original function.

y=b^x:
Domain: Range:
0 -----> 1
1 -----> 4
2 -----> 16
log_b(x)=y
Domain: Range:
1 -----> 0
4 -----> 1
16 -----> 2
See how the logarithmic function's domain is the exponential function's range and the logarithmic function's range is the exponential function's domain. In a simpler way of thinking about it, you swap the domain and range, or x and y. This is also why with a function with the point (x,y), the inverse of the function will have the point (y,x).
(1 vote)
• Hi, can you please explain how did you know that y=log_b(x) is inverse of y=b^x. a detailed answer would be appreciated. Thanks.
• The actual definition of a logarithm that proves this is several years ahead of the math you're having now, so it wouldn't mean much to you at this stage.

But, the short answer is that the logarithm was invented to be the inverse operation of an exponential function. In other words, there is a difficult mathematical process that was worked out over the course of about 80 years by Nicholas Mercator, Leonhard Euler and others. The term "log" is just a shorthand way of expressing this difficult bit of math.

So, a log is just a quick way of writing the inverse function of an exponential function. Thus, by definition, the log must be the inverse function of the exponential function.

So, the reason you are not given exactly what a log is or how to compute it by hand at this level of study is that it is very difficult math. Since we have calculators that can do this math for us, it is possible for people to use logs without having to master very advanced and difficult computations.

But, just for reference, here is one way to write the actual definition of a logarithm
log_b (x)= lim n→0 [x^(n) - 1] / [b^(n) - 1]
where b is the base of the logarithm.
• At , how do we know that b raised to the power 1 is equal to four. Is it a definite variable with its value
• It is a constant that we deduced from the graph that is given to us.
• I'm sorry but I'm confused by this video. I've understood everything so far but I got lost. How does this work?
• You have your input x in the Log function, and the output is graphed.
Like when x=2 on the equation and say it equals 5 when x=2
you would graph 2,5
• To solve for other points, couldn't I figure out that b=4 from the pattern in y=b^x and go from there?
``20 log (0.1/2)``
``20 * log_10 (0.1/2)``