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Evaluating logarithms: change of base rule

Sal approximates log₅(100) by rewriting it as log(100)/log(5) using the change of base rule, then evaluates with a calculator. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • leaf green style avatar for user Daryl
    I posted an answer to this question with an alternate proof to the change of base formula that I find more intuitive. I did this because neither the comment nor question sections have enough space for it.
    (24 votes)
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    • leaf green style avatar for user Daryl
      Let logₐ(b) = s. Then we wish to show that, for any value of x, logₐ(b) = logₓ(b) / logₓ(a). Let's say logₓ(a) = t and logₓ(b) = u.


      Change each of these to the exponent notation:

      logₐ(b) = s means aˢ = b.

      logₓ(a) = t means xᵗ = a.

      logₓ(b) = u means xᵘ = b.

      So we know that aˢ = xᵘ, because they both equal b. We also know a = xᵗ now, so we can substitute xᵗ for a in the equation aˢ = xᵘ, and get (xᵗ)ˢ = xᵘ. This simplifies to xˢᵗ = xᵘ using the exponent properties, which implies st = u, or s = u/t. Recall that logₐ(b) = s, logₓ(a) = t, and logₓ(b) = u. So we've now shown with s = u/t that logₐ(b) = logₓ(b) / logₓ(a). This completes the proof.
      (60 votes)
  • leaf grey style avatar for user cmaryk12296
    Where can I download a graphing calculator for my computer like what Sal is using?
    (12 votes)
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  • spunky sam blue style avatar for user Chunmun
    So why do we always take log(base 10) ?
    Why we dont take any other number ?
    (11 votes)
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    • leaf orange style avatar for user TB
      You can do that. In fact base 10, base 2 and base e (natural logarithm) are only the most common.
      The limitation is your calculator. Simple calculators only have a function for base 10 and base e and that's it.
      But if you had log[base 5] of 125, you could do it in your head. The answer being 3.
      (21 votes)
  • blobby green style avatar for user mdimen
    I didn't get that last part of the video, where Clog(a) = log(b) turns into the very last part of what Khan was writing about, logb/loga. How does that work out? Please and thank you :)
    (9 votes)
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  • blobby blue style avatar for user :D
    this makes no sense :C
    (7 votes)
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    • female robot amelia style avatar for user him
      Certainly! Let's go through the explanation of the change of base rule for evaluating logarithms:

      *Change of Base Rule for Logarithms*:

      The change of base rule is a property of logarithms that allows us to rewrite the logarithm of any base as the logarithm of another base. Specifically, for any positive real numbers "a," "b," and "c," where "a" and "b" are the bases of the logarithms and "c" is the argument, we have:

      logₐ(c) = log_b(c) / log_b(a)

      *Explanation*:

      Suppose we want to evaluate a logarithm with a certain base "a," but our calculator only has a logarithm function with a different base "b." The change of base rule provides a way to convert the logarithm with base "a" into an equivalent logarithm with base "b."

      Let's use an example to illustrate the process:

      *Example*:
      Evaluate log₅(100) using a calculator that only has a logarithm function with base 10.

      *Step 1*:
      Apply the change of base rule:
      log₅(100) = log(100) / log(5)

      *Step 2*:
      Evaluate the logarithms on the right-hand side using the calculator with base 10 logarithm (log) function:
      log(100) ≈ 2 (since 10^2 = 100)
      log(5) ≈ 0.69897 (approximately)

      *Step 3*:
      Substitute the values back into the expression:
      log₅(100) ≈ log(100) / log(5) ≈ 2 / 0.69897 ≈ 2.86135

      *Result*:
      The approximation of log₅(100) using the change of base rule and a calculator with base 10 logarithm function is approximately 2.86135.

      By using the change of base rule, we can evaluate logarithms with different bases using a calculator that supports a specific base for logarithms. This rule provides a helpful technique for logarithmic calculations when using calculators or software that may not have specific logarithm functions for every base.

      -----------------------------------------------------
      The change of base rule is a useful logarithmic property that allows us to evaluate logarithms of any base by using a calculator with a different base logarithm function. This property states that for any positive real numbers "a," "b," and "c," where "a" and "b" are the bases of the logarithms and "c" is the argument:

      logₐ(c) = log_b(c) / log_b(a)

      In the given example, Sal wants to approximate log₅(100) using a calculator that has a logarithm function with base 10 (log), but not with base 5 (log₅). So, Sal uses the change of base rule to convert log₅(100) into an expression that can be evaluated using the calculator:

      log₅(100) = log(100) / log(5)

      Now, Sal can use a regular calculator to find the logarithms of 100 and 5, which are:

      log(100) ≈ 2 (since 10^2 = 100)
      log(5) ≈ 0.69897 (approximately)

      Now, substitute these values back into the expression:

      log₅(100) ≈ log(100) / log(5) ≈ 2 / 0.69897 ≈ 2.86135

      So, the approximation of log₅(100) using the change of base rule and a calculator with base 10 logarithm function is approximately 2.86135.
      (6 votes)
  • mr pants teal style avatar for user Kairi
    How can you do this WITHOUT a calculator?
    (5 votes)
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  • piceratops ultimate style avatar for user MusabTheBoss
    Can you put any number as a base in the logarithmic base conversion formula?
    (6 votes)
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  • duskpin sapling style avatar for user Camila Murphy
    How would i solve 3^(x − 4) = 6 using the change of base formula.
    (5 votes)
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    • female robot grace style avatar for user loumast17
      3^(x-4) = 6

      (3^x)/(3^4) = 6 by the exponent properties

      3^x = 6*81 multiply both sides by 3^4 or 81

      log_3(486)=x by converting exponent for to log form. Thsi reads as log base 3 of 486

      Fromt here you can make it log(486)/log(3) where the logs are any base you want.
      (5 votes)
  • leafers tree style avatar for user Vishwa Patel
    How can I evaluate “log10 5” without a calculator? I know there has to be a way.

    Also, why can’t (log10 100)/(log10 5) be written as
    “log10 100 - log10 5”? The answer to expression is 1.3010.
    (4 votes)
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    • blobby green style avatar for user habrother01
      We know that log_10 1 = 0 and log_10 10 = 1. From this, we can deduce this: 0 < log_10 5 < 1. We can specify more from this because log(√10) = 1/2. This means
      1/2 < log_ 10 5) < 1 because √10 < 5. Using this method, you would be able to get close to log_10 5, but it would be very tedious work. Also,
      (log_10 100)/(log_10 5) is something very different from log_10 100 - log_10 5 because if we say
      A = log_10 100 and
      B = log_10 5, we can come up with the following equations:
      log_10 100 / log_10 5 = A / B, and
      log_10 100 - log_10 5 = A - B.
      These two equations can't ever be equal to each other unless A and B are both zeros.
      (4 votes)
  • leafers sapling style avatar for user maxuphigh
    For the Change of base formula, can (x) be anything you desire or does it HAVE to be 10?
    (3 votes)
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    • male robot donald style avatar for user Venkata
      It can be anything. You can change it based on the question asked.

      There's also a neat way to prove that the base can be anything. Here it is, if you're interested:

      Let log_a(b) = y (log_a(b) is just log of b with base a)

      So, from this, we have a^y = b.

      Taking log_d on both sides, we have

      log_d(a^y) = log_d(b)

      Using log properties, we get

      y * log_d(a) = log_d(b)

      y = (log_d(b))/(log_d(a))

      log_a(b) = (log_d(b))/(log_d(a))

      Here, a, b and d are all positive reals greater than 1
      (4 votes)

Video transcript

Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base. And that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense, or how we can derive it. So the change of base formula just tells us that log-- let me do some colors here-- log base a of b is the exact same thing as log base x, where x is an arbitrary base of b, over log base, that same base, base x over a. And the reason why this is useful is that we can change our base. Here are our bases, a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying-- this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x, and e is obviously the number 2.71, keeps going on and on and on forever. Now let's apply it to this problem. We need to figure out the logarithm-- and I'll use colors-- base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing as log-- I'll make x 10-- log base 10 of 100 divided by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100-- what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2. So it simplifies to 2 over log base 10 of 5. And we can now use our calculator, because the log function on a calculator is log base 10. So let's get our calculator out. We want to clear this. 2 divided by-- When someone just writes log, they mean base 10. If they press LN, that means base e. So log without any other information is log base 10. So this is log base 10 of 5 is equal to 2 point-- and they want it to the nearest thousandth-- so 2.861. So this is approximately equal to 2.861. And we can verify it because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power. And this number is closer to 3 than it is to 2. Well, let's verify it. So if I take 5 to that power, and then let me type in-- let me just type in what we did to the nearest thousandth-- 5 to the 2.861. So I'm not putting in all of the digits. What do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good. That this is the power that I have to raise 5 to to get to 100. Now with that out of the way, let's actually think about why this property, why this thing right over here makes sense. So if I write log base a-- I'll try to be fair to the colors-- log base a of b. Let's say I set that to be equal to some number. Let's call that equal to c, or I could call it e for-- Well, I'll say that's equal to c. So that means that a to the c-th power is equal to b. This is an exponential way of writing this truth. This is a logarithmic way of writing this truth. This is equal to b. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this? 10 to the same power will be equal to this, because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base. And I'll actually do log base x to prove the general case, here. So I'm going to take log of base x of both sides of this. So this is log base x of a to the c power-- I try to be faithful to the colors-- is equal to log base x of b. And let me close it off with orange, as well. And we know from our logarithm properties, log of a to the c is the same thing as c times the logarithm of whatever base we are of a. And of course, this is going to be equal to log base x of b. Let me put-- I can just write a b, right over there. And if we wanted to solve for c, you just divide both sides by log base x of a. So you would get c is equal to-- and I'll stick to the color-- so it's log base x of b, which is this, over log base x of a. And this was what c was. c was log base a of b. It's equal to log base a of b. Let me write it this way. Let me write it-- Well, let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going, but I want to be fair to the colors. So c is equal to log base x of b over-- let me scroll down a little bit-- log base x, dividing both sides by that, of a. And we know from here I can just copy and paste it, this is also equal to c. This is how we defined it. So let me copy it and then let me paste it. So this is also equal to c. And we're done. We've proven the change of base formula. Log base a of b is equal to log base x of b divided by log base x of a. In this example, a was 5, b is 100, and the base that we switched it to is 10. x is 10.