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## Algebra (all content)

### Unit 11: Lesson 27

Solving exponential equations with logarithms (Algebra 2 level)

# Solving exponential equations using logarithms: base-10

Sal solves the equation 10^(2t-3)=7. Created by Sal Khan.

## Want to join the conversation?

• Couldn't we just oversimplify it to (log 7000)/(log 100) ?
• Yes, this can be done. We can see that (log 7000)/(log 100) is equivalent to the correct answer given, which is [(log 7) + 3]/2, using definitions and laws of logarithms:

(log 7000)/(log 100) = [log(7*10^3)] / [log (10^2)] = [(log 7) + log (10^3)] / [log (10^2)]
= [(log 7) + 3]/2.

Have a blessed, wonderful day!
• I am having trouble solving 3log(4x+3) < 1. When I computed the problem I got x > 5/6.
I 1st divided both sides by 3, then set it up as e^4x + e^3 < e^1/3. which i then rewrote as 4x+3 < 1
I don't think my answer is correct but I am lost as to where I went wrong.
Pease help!!
• Dividing by 3 is the correct 1st step, but from there, I like to convert it back to its exponential form. That means that if this is a "common logarithm" (base 10), you would write 4x+3 < 10^(1/3). This can also be attained by raising both sides to a power of 10 [i.e. 10^(log(4x+3)) < 10^(1/3)]. The 10^log cancels itself out and you are left with the 4x+3 on the left. Solving 4x+3 < 10^(1/3) leaves us with x < [-3 + 10^(1/3)] / 4 or approximately x < -0.2114. If you plug it into the original inequality, you can verify the correct answer.
• I am having trouble with solving a exponential equation using a logarithm. The equation is 5(10)^x-31=81.6 , I keep trying every way to solve this but it's just not giving me the correct answer. Help!
• 1st Isolate the base with the exponent by dividing both sides by 5 and you get:
10^x-31=16.32

2nd log both sides

log 10 of 10^x-31=log 10 of 16.32
The log 10 and 10 cancel out, your left with:
x-31=log 10 of 16.32

3rd add 31 to both sides to isolate x
x=log 10 of 16.32 +31

4th Depending on your calculator, you will either press the log button first and then enter the value or you will enter the value first and then press the log button.
For the windows calculator: Type in 16.32 and then click the log button and then click on enter.
You get 1.2127
Then add 31 to that value: 32.213
( I rounded to the nearest thousandth)
Check to see if answer is correct:
Plug 32.213 into x
10^32.213=1.63305...
1.63305*5=8.165259...
Subtract 31 from that number
8.165259
So yes, our answer is correct
• 7=x^2.8073 how do you solve this with logs
• You don't need logs for this. Just raise both sides to the 1/2.8073 power and then you will have your answer.
• Is there a video for an e-base problem instead of a base with 10?
• What happens if you have a power on either side, for example: 4^2x+3 = 5^3x-1
• Take logs on both sides. log(4^2x+3) = log(5^3x-1)
Then use the log rules to bring down the power 2x+3log4 = 3x-1log5
You can then split these logs up 2xlog4 + 3log4 = 3xlog5 - log5
Get all your x values over to one side 3xlog5 - 2xlog4 = 3log4 + log5
Then factorise to take out x x (3log5 - 2log4) = 3log4 + log 5
And rearrange to find x x = (3log4 + log5) / (3log5 - 2log4)
• So I'm studying for a test. In one of my practice problems, I have to solve for x in the equation 12^(x-3)=17^(2x), and I am supposed to write the exact answer in terms of log_10. How do I solve it?

One other thing I'm curious about: why does solving for a variable in terms of log_10 work?
• Since we're dealing with variables as indices (aka powers, exponents), log_10, log_(anything), and ln (the natural log) all work because they are inverses of exponential functions. By using log, you can easily manipulate the exponents making it easier to solve.
``12^(x - 3) = 17^(2x) {Equation from Question}log(12^(x - 3)) = log(17^(2x)) {Log both sides} I'll be referring to log_10 as log from now on.(x - 3)log(12) = 2x(log(17)) {Standard rule with logarithms, essential for solving these types of questions}xlog(12) - 3log(12) = 2x(log(17)) {Distributing on Left Hand Side}xlog(12) - 2x(log(17)) = 3log(12) {Reorganising the Equation}x(log(12) - 2log(17)) = 3log(12) {Factoring Out 'x'}x = 3log(12)/(log(12) - 2log(17)) {Isolate 'x'}``

And I think I answered your second question on top. Hope this helps. :-)
(1 vote)
• "-(-2)^n = 1532" how do I figure out the value of n?
• open up the brackets and you will get positive 2.
now as far my brain works. you can factor out 1532.
this would be a lengthy task so you can apply a shortcut. we know 2^4=16
so divide 1532 by 32 and you get let's say x.
then multiply x by 4 and here you get the answer.
in case there is a remaider factor out the remainder and add the result to x*4.
hope it helps.