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.

## College Algebra

### Course: College Algebra>Unit 14

Lesson 4: Exponential model word problems

# 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 $a\cdot b^x=d$a, dot, b, start superscript, x, end superscript, equals, d

Let's solve 5, dot, 2, start superscript, x, end superscript, equals, 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!
\begin{aligned} 5\cdot 2^x&=240 \\\\ 2^x&=48 \end{aligned}
Now, we can solve for x by converting the equation to logarithmic form.
start color #11accd, 2, end color #11accd, start superscript, start color #1fab54, x, end color #1fab54, end superscript, equals, start color #e07d10, 48, end color #e07d10 is equivalent to log, start base, start color #11accd, 2, end color #11accd, end base, left parenthesis, start color #e07d10, 48, end color #e07d10, right parenthesis, equals, start color #1fab54, x, end color #1fab54.
And just like that we have solved the equation! The exact solution is x, equals, log, start base, 2, end base, left parenthesis, 48, right parenthesis.
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.
\begin{aligned} x &= \log_{2}(48) \\\\ &=\dfrac{ \log(48)}{\log(2)} &&{\gray{\text{Change of base rule}}} \\\\ &\approx 5.585 &&{\gray{\text{Evaluate using calculator}}} \end{aligned}
The approximate solution, rounded to the nearest thousandth, is x, approximately equals, 5, point, 585.

1) What is the solution of 2, dot, 6, start superscript, x, end superscript, equals, 236?

2) Solve 5, dot, 3, start superscript, t, end superscript, equals, 20.
t, equals

3) Solve 6, dot, e, start superscript, y, end superscript, equals, 300.
y, equals

## Solving exponential equations of the form $a\cdot b^{cx}=d$a, dot, b, start superscript, c, x, end superscript, equals, d

Let's take a look at another example. Let's solve 6, dot, 10, start superscript, 2, x, end superscript, equals, 48
We start again by isolating the exponential part by dividing both sides by 6.
\begin{aligned} 6\cdot 10^{2x}&=48\\\\ \blueD{10}^{\greenD{2x}}&= \goldD8 \end{aligned}
Next, we can bring down the exponent by converting to logarithmic form.
\begin{aligned} \log_{\blueD{10}}(\goldD8)&=\greenD{2x} \end{aligned}
Finally, we can divide both sides by 2 to solve for x.
x, equals, start fraction, space, log, start base, 10, end base, left parenthesis, 8, right parenthesis, divided by, 2, end fraction
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.
\begin{aligned} x&=\dfrac{~{\log_{10}(8)}}{2} \\\\ &= \dfrac{~{\log(8)}}{2}&&{\gray{\log_{10}(x)=\log(x)}} \\\\ &\approx 0.452 &&{\gray{\text{Evaluate using calculator}}}\end{aligned}

4) Which of the following is the solution of 3, dot, 10, start superscript, 4, t, end superscript, equals, 522?

5) Solve 4, dot, 5, start superscript, 2, x, end superscript, equals, 300.
x, equals

6) Solve minus, 2, dot, 3, start superscript, 0, point, 2, z, end superscript, equals, minus, 400.
z, equals

## Challenge problem

7) Which of the following are solutions to left parenthesis, 2, start superscript, x, end superscript, minus, 3, right parenthesis, left parenthesis, 2, start superscript, x, end superscript, minus, 4, right parenthesis, equals, 0?

## 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.
• 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
• 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.
• 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.
• Can anybody help:
log_5 y+log_y 5= 6 ?
• 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.
• 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.
• 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 = 02^x = 42 * 2 = 4 x = 2 as the other answer.

This is how it is sorted out. If you still have questions, please comment :)
• What exactly is going on when you have to change the base rule? I don't really follow that part of these equations
• How do you do the following problem log4n=1.5log416+1.3log464
• What if our equation has variables within the exponent. My equation looks like this, (1/9)^(2x+1) (27^x-2)=1
• You need to get both sides to have a common base.
1) Multiply both sides by 9^(2x+1). This changes your equation into: 27^(x-2)=9^(2x+1)
2) Factor 27 and 9. 27=3^3 and 9 = 3^2. So, 3 can be used as a common base. Replace 27 and 9 with their factors:
(3^3)^(x-2) = (3^2)^(2x+1)
3) Multiply the exponents: 3^(3x-6) = 3^(4x+2)

Now, for these 2 sides to be equal, the exponents must be equal. So, you can solve: 3x-6=4x+3 to find x.

Hope this helps.
• i really need help on this on because in did not even learn algebra so it would really help if someone could hel me pls and thank you🙂🙂🙂🙂🙂🙂🙂
• What if the question was 4^(x+1)=30. How would you place '1' in the logarithm?
(1 vote)
• It's the same thing, just solve for x+1 before you solve for x.

4^(x+1) = 30
log₄(4^(x+1)) = log₄(30)
x+1 = log₄(30)
x = log₄(30) - 1

x ≈ 1.45344

2x, x+1, 19236x-31341742, πx^2+6x+2, you can really replace any function with "x" in this equation and then solve for that.
• how to solve 10^2z/3=15
(1 vote)
• 10^(2𝑧)∕3 = 15

10^(2𝑧) = 45

2𝑧 = log 45

𝑧 = log(45)∕2

– – –

10^(2𝑧∕3) = 15

2𝑧∕3 = log 15

𝑧 = 3 log(15)∕2