If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Solving exponential equations using logarithms

Learn how to solve any exponential equation of the form a⋅b^(cx)=d. For example, solve 6⋅10^(2x)=48.
The key to solving exponential equations lies in logarithms! Let's take a closer look by working through some examples.

Solving exponential equations of the form abx=d

Let's solve 52x=240.
To solve for x, we must first isolate the exponential part. To do this, divide both sides by 5 as shown below. We do not multiply the 5 and the 2 as this goes against the order of operations!
52x=2402x=48
Now, we can solve for x by converting the equation to logarithmic form.
2x=48 is equivalent to log2(48)=x.
And just like that we have solved the equation! The exact solution is x=log2(48).
Since 48 is not a rational power of 2, we must use the change of base rule and our calculators to evaluate the logarithm. This is shown below.
x=log2(48)=log(48)log(2)Change of base rule5.585Evaluate using calculator
The approximate solution, rounded to the nearest thousandth, is x5.585.

Check your understanding

1) What is the solution of 26x=236?
Choose 1 answer:

2) Solve 53t=20.
Round your answer to the nearest thousandth.
t=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

3) Solve 6ey=300.
Round your answer to the nearest thousandth.
y=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Solving exponential equations of the form abcx=d

Let's take a look at another example. Let's solve 6102x=48
We start again by isolating the exponential part by dividing both sides by 6.
6102x=48102x=8
Next, we can bring down the exponent by converting to logarithmic form.
log10(8)=2x
Finally, we can divide both sides by 2 to solve for x.
x= log10(8)2
This is the exact answer. To approximate the answer to the nearest thousandth, we can type this directly into the calculator. Notice here that there is no need to change the base since it is already in base 10.
x= log10(8)2= log(8)2log10(x)=log(x)0.452Evaluate using calculator

Check your understanding

4) Which of the following is the solution of 3104t=522?
Choose 1 answer:

5) Solve 452x=300.
Round your answer to the nearest thousandth.
x=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

6) Solve 230.2z=400.
Round your answer to the nearest thousandth.
z=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Challenge problem

7) Which of the following are solutions to (2x3)(2x4)=0?
Choose all answers that apply:

Want to join the conversation?

  • mr pants purple style avatar for user Cookenmaster, Cheyenne
    In number 5, where did the 1/2 come from. Why... Actually, it would be great if the whole problem is explained to me, but differently. It doesn't make sense to me.
    (16 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user InnocentRealist
      I didn't use "1/2" and I don't think it's really necessary:

      1) 4*(5^(2*x)) = 300

      2) 5^(2*x) = 75 (At this point you can just take the log of both sides (see below) and that's where the 1/2 comes from), or:

      3) (5^2)^x = 75 (laws of exponents)

      4) 25^x = 75

      5) x = log_25(75) (log form)

      6) x = log75/log25 = 1.341 (change of base rule)

      Or for 3) you could do:

      3) log_5(5^(2*x)) = log_5(75) (log form)

      4) 2*x = log_5(75) (Here's where 1/2 comes in)

      5) x = (1/2)log_5(75)

      6) x = (1/2)(log75/log5) = 1.341
      (24 votes)
  • piceratops tree style avatar for user aditya
    while practicing a question came to my mind. It maybe stupid,but I am going to ask it anyway. we have a subtraction property(log a-log b=log a/b) and change of base property(log_a b=log b/log a). I can understand when it goes in the said direction.but when it is applied in reverse how do i differentiate eg. log a/log b how so i determine which property to apply.
    (7 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user Katriana
      Your question was a long time ago, but here is an answer for anyone that may be wondering.
      The subtraction property says that log(a)-log(b) is equal to log(a/b).
      The change of base property says that log_a(b) is equal to (log_x(b))/(log_x(a)).
      So, in the subtraction property the division is within the log, while for the change of base property we are really dividing the answers from the two logs.
      So, if we are given log(a/b) to expand, we can use the subtraction property; if we are to condense (log(b))/(log(a)), then we should use the change of base property.
      I had a little difficulty understanding how to tell the difference myself at first. Hopefully this explanation was written clearly enough.
      (10 votes)
  • old spice man blue style avatar for user Ruzbah Aria
    Can anybody help:
    log_5 y+log_y 5= 6 ?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • duskpin sapling style avatar for user Claire Z
    cant wait to come back years from now and realize i still cant do algebra 2
    (9 votes)
    Default Khan Academy avatar avatar for user
  • mr pink green style avatar for user b.k.phillips
    Can someone walk me through the last challenge question? I see solving for each zero as they suggested, but am curious if one can expand it out as a quadratic to solve. When I try to do this, I keep coming out w incorrect solutions.

    CHALLENGE QUESTION: Which of the following are solutions to (2^x-3)(2^x-4)=0

    My approach is giving me {WHEN SETTING 2^x =y}:
    y^2-7y+12=0

    which then gives me a quadratic formula of;
    7+or- sqrt(49-4(12)) all over 2

    which results in values of 4 and 3.

    I assume I am making an error, perhaps in setting the y or somewhere else regarding my treatment in expanding the equation, but I am not finding it; any help is appreciated.
    (4 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Duymayan, Beyza
      In this setting we are solving for x, and how nice of them, to already give it to us factored out. When this quadratics is factored out this way, it means we are 1 step away from finding x itself. Both set of equations equal to 0 since we separated the into two individual parts.

      2^x - 3 = 0
      2^x = 3
      x = log base 2 parentheses 3 is one of the answers.

      2^x - 4 = 0
      2^x = 4
      2 * 2 = 4
      x = 2 as the other answer.


      This is how it is sorted out. If you still have questions, please comment :)
      (7 votes)
  • duskpin seed style avatar for user DaySatyr5771669
    Hi I was wondering if you could help me with this question?
    "Find the value of x in terms of a, where 3 log(base a)x=3+log(base a) 8"
    Thanks
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Edward Miotke
      I believe that the short answer is: x = 2a.

      Below is an explanation of how I got to that answer:

      Start with: 3log_a (x) = 3 + log_a (8)
      Multiply both sides by 1/3: log_a (x) = 1 + (1/3)log_a (8)
      After using the power rule we get: log_a (x) = 1 + log_a (8^(1/3))
      which simplifies to: log_a (x) = 1 + log_a (2)

      Since 1 = log_a (a), we can rewrite the equation as:
      log_a (x) = log_a (a) + log_a (2)
      Next, using the Product Rule we get: log_a (x) = log_a (2a)

      Finally, after canceling the log from both sides we get: x = 2a.

      I hope that this is helpful.
      (5 votes)
  • blobby green style avatar for user ys1006679
    log down 4 (2x/x-1)=2
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Amy Shaffer
    Zero product rule does not yield same result as quadratic rule from unit 2 lesson 6. Quadratic yielded (17+- 1)/12 = 1.5 or 1.3333. Explain.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user george johnston
    What exactly is going on when you have to change the base rule? I don't really follow that part of these equations
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user trevor pummell
    How do you do the following problem log4n=1.5log416+1.3log464
    (2 votes)
    Default Khan Academy avatar avatar for user