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### Course: Algebra 2 > Unit 8

Lesson 5: Solving exponential equations with logarithms- Solving exponential equations using logarithms: base-10
- Solving exponential equations using logarithms
- Solve exponential equations using logarithms: base-10 and base-e
- Solving exponential equations using logarithms: base-2
- Solve exponential equations using logarithms: base-2 and other bases

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# 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 $a\cdot {b}^{x}=d$

Let's solve $5\cdot {2}^{x}=240$ .

To solve for $x$ , we must first isolate the exponential part. To do this, divide both sides by $5$ as shown below. We do $5$ and the $2$ as this goes against the order of operations!

**not**multiply theNow, we can solve for $x$ by converting the equation to logarithmic form.

And just like that we have solved the equation! The $x={\mathrm{log}}_{2}(48)$ .

**exact**solution isSince $48$ 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.

The $x\approx 5.585$ .

**approximate**solution, rounded to the nearest thousandth, is### Check your understanding

## Solving exponential equations of the form $a\cdot {b}^{cx}=d$

Let's take a look at another example. Let's solve $6\cdot {10}^{2x}=48$

We start again by isolating the exponential part by dividing both sides by $6$ .

Next, we can bring down the exponent by converting to logarithmic form.

Finally, we can divide both sides by $2$ to solve for $x$ .

This is the $10$ .

**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### Check your understanding

## Challenge problem

## Want to join the conversation?

- 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.(18 votes)
- 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(26 votes)

- 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.(8 votes)
- 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.(11 votes)

- cant wait to come back years from now and realize i still cant do algebra 2(11 votes)
- Can anybody help:

log_5 y+log_y 5= 6 ?(7 votes)- Really nice question! You need to know the properties of logarithms in order to solve this problem.

It would be too tedious to type out the answer for you, so instead I linked it in this graph step by step.

https://www.desmos.com/calculator/9teeitvqjp

The answer is y = x=5^(3-2√2)

Hope this helps.(9 votes)

- 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)- 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)

- 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)- 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)

- log down 4 (2x/x-1)=2(2 votes)
- Switch the equation into exponential form to solve:

2x/x-1 = 4^2

Simplify the exponent: 2x/x-1 = 16

Then, assuming your expression inside the parentheses is (2x)/(x-1), you can cross multiply to get: 2x=16(x-1), and solve for "x".

Hope this helps.(6 votes)

- 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)
- This questions doesn't apply to to this lesson. Please post your question in the lesson you are asking about. Without knowing the actual original problem and the answers derived for that problem, it is impossible to give you an answer.(7 votes)

- 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)
- Did you watch the last section of videos "The Change of Base Formula for Logarithms"? That really helped me, since I'd never heard of changing bases before. Here's a link to the first video:

https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/change-of-base-formula-for-logarithms/v/change-of-base-formula(3 votes)

- How do you do the following problem log4n=1.5log416+1.3log464(2 votes)
- Assuming this is the problem:

log_4(n) = 1.5log_4(16) + 1.3log_4(64)

1) log_4(16) = 2 and log_4(64) = 3

So we have : log_4(n) = 1.5(2) + 1.3(3)

2) log_4(n) = 6.9

This is equivalent to: 4^(6.9) = n

3) 14,263.1 ≈ n(1 vote)