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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.

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  • leaf green style avatar for user Alex Wong
    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
    (168 votes)
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    • blobby green style avatar for user jaditya23
      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.
      (18 votes)
  • blobby green style avatar for user Sara
    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.
    (45 votes)
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    • leaf green style avatar for user johnmgiorgi
      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!
      (43 votes)
  • leaf green style avatar for user Ghoshofx1
    at around , why is sal taking the y values of the first table and using them as the x values of the second table?
    (11 votes)
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    • duskpin ultimate style avatar for user hasitak11170
      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.
      (9 votes)
  • aqualine ultimate style avatar for user Kaya
    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!
    (6 votes)
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    • aqualine tree style avatar for user Judith Gibson
      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.)
      (9 votes)
  • leafers tree style avatar for user Tatewisaacs03
    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?
    (10 votes)
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  • duskpin seed style avatar for user kofiamosah1
    At , how do we know that b raised to the power 1 is equal to four. Is it a definite variable with its value
    (7 votes)
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  • leaf yellow style avatar for user Adeeb Mohammed Islam
    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.
    (5 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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.
      (8 votes)
  • hopper jumping style avatar for user :)
    I'm sorry but I'm confused by this video. I've understood everything so far but I got lost. How does this work?
    (7 votes)
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  • piceratops ultimate style avatar for user Onetwothree45
    To solve for other points, couldn't I figure out that b=4 from the pattern in y=b^x and go from there?
    (5 votes)
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    • piceratops ultimate style avatar for user James Randall
      From this set it is obvious that b=4, but the point of the exercise is to highlight the fact that the logarithmic function is an inverse of the exponential function, meaning that you still could have mapped the second set of points simply by reflecting over the line y = x, even if the values present did not provide a simple/obvious value for b.
      (5 votes)
  • blobby green style avatar for user Beth McEwen
    I've seen a notation such as 10 log (25/5) = 10 log (5). I'm confused about the "10" and the "log" placement. Is this the same as saying log_10 5? Another (very similar) notation is 20 log (0.1/2)=20 log (0.05). Again, why put the 10 and the 20 before the "log"? Thank You!
    (1 vote)
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Video transcript

Voiceover:The three points plotted below are on the graph of y is equal to b to the x power. Based only on these three points, plot the three corresponding points that must be on the graph of y is equal to log base b of x by clicking on the graph. I've actually copy and pasted this problem on my little scratch pad so I can mark it up a little bit. What is this first function? This first function is telling us so x, and this is y is equal to b to the x power. When x is equal to zero, y is equal to one. That's this point right over here. When x is equal to one, b to the one power or b to the first power is equal to four. y is equal to four. Another way of thinking of this y or four is equal to b to the first power and actually we can do [said] must be four. That's this point right over there. This point is telling us that b to the second power is equal to 16. When x is equal to two, b to the second power, y is equal to 16. Now we want to plot the three corresponding points on this function. Let me draw another table here. Now it's essentially the inverse function where this is going to be x and we want to calculate y is equal to log base b of x. What are the possibilities here? What I want to do is think ... Let's take these values because these are essentially inverse functions log is the inverse of exponents. If we take the points one, four, and 16. What is y going to be here? y is going to be log base b of one. This is saying what power I need to raise b to to get to one. If we assume that b is non zero and that's a reasonable assumption because b to different powers are non zero, this is going to be zero for any non zero b. This is going to be zero right there, over here. We have the point one comma zero, so it's that point over there. Notice this point corresponds to this point, we have essentially swapped the x's and y's. In general when you're taking an inverse you're going to reflect over the line, y is equal to x and this is clearly reflection over that line. Now let's look over here, when x is equal to four what is log base b of four. What is the power I need to raise b to to get to four. We see right over here, b to the first power is equal to four. We already figured that out, when I take b to the first power is equal to four. This right over here is going to be equal to one. When x is equal to four, y is equal to one. Notice once again, it is a reflection over the line y is equal to x. When x is equal to 16 then y is equal to log base b of 16. The power I need to raise b to, to get to 16. Well we already know, if we take b squared, we get to 16, so this is equal to two. When x is equal to 16, y is equal to two. Notice we essentially just swapped the x and y values for each of these points. This is y, this is a reflection over the line y is equal to x. Now, let's actually do that on the actual interface. The whole reason is to give you this appreciation that these are inverse functions of each other. Let's plot the points. That point corresponded to that point, so x zero, y one corresponds to x one, y zero. Here x is one, y is four that corresponds to x four, y one. Here x is two, y is 16 that corresponds to x 16, y is two. We got it right.