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## Integrated math 3

### Course: Integrated math 3 > Unit 5

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: 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) ?(6 votes)
- 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!(14 votes)

- 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!!(3 votes)- 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.(8 votes)

- 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!(4 votes)
- 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(3 votes)

- 7=x^2.8073 how do you solve this with logs(3 votes)
- You don't need logs for this. Just raise both sides to the 1/2.8073 power and then you will have your answer.(5 votes)

- Why is 10 specifically used as the common logarithm base?(2 votes)
- Our number system is base 10. So, it makes sense that the most commonly used log would also be base 10.(5 votes)

- I keep getting problems wrong because I evaluate too precisely. This might be a basic question, but how do I know how precise to be? When evaluating logs with the change of base I generally evaluate each log to the fourth decimal place (ten-thousandths), then after combining terms I round to the nearest thousandth. For instance, if evaluating log8/log5 I would use the calculator to find 0.9030/0.6989 = 1.2920, which is 1.292 when rounded to the thousandths place. (That wasn't the best example, but I hope that made sense.) Khan, though gives a different number for the thousandths place, making me wonder how I am going wrong. Please help!(3 votes)
- when doing math problems, it is best to not round until you reach the final answer. if you are using a calculator to find the logs you used in the change of base formula, you can simply use the fraction function and then type in the logs to find the answer, rather than taking a rounded number of each and calculating with them.

example:

0.0012999/0.0023999=0.541647568...=0.542

rounding this would give you

0.001/0.002=0.5.

(the example is over exaggerated, but you get the point.)(2 votes)

- Is there a video for an e-base problem instead of a base with 10?(3 votes)
- The lesson here (https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs#x2ec2f6f830c9fb89:e) covers logarithms with a base of e (also known as a natural log).(2 votes)

- What happens if you have a power on either side, for example: 4^2x+3 = 5^3x-1(2 votes)
- 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)(4 votes)

- wait...so what's the difference between base 10 and base -10?(2 votes)
- "base 10" and "base-10" are the same. Typically, math books will include a hyphen between "base" and the number.(3 votes)

- 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?(3 votes)- 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)

## Video transcript

Voiceover:Solve the
equation for T and express your answer in terms
of base 10 logarithms. And this equation is 10 to
the 2T - 3 is equal to 7. We want to solve for T in
terms of base 10 logarithms. So let me get my little
scratchpad out and I've copied and pasted the same problem. So I'm just going to
rewrite it, so they have 10 to the 2T-3 is equal to 7. Actually, let me color
code this a little bit. So 10 to the 2T - 3 is equal to 7. So this is clearly an
exponential form right over here. if we want to write it
in logarithmic form, where we could, that'll
essentially allow us to solve for the exponent, so we could say, this is the exact same
truth about the universe as saying that the log base
10 of 7 is equal to 2T - 3. Let's just make sure that makes sense, this is saying 10 to the 2T - 3 = 7 This is saying that the power
that I need to raise 10 to to get to 7 is 2T - 3, or 10
to 2T - 3 power is equal to 7. So these are equivalent statements. What this form does is it
starts to put it into a form that's easier to solve for
T, now if we want to solve for T, we can add 3 to both sides. So if we add 3 to both
sides, we are going to get, log base 10 of 7 + 3, plus 3. And this +3 is of course
outside of the logarithm to make it clear, it's just like that. And on the right hand side
this is going to be equal to, this is going to be equal to just 2T. And now to solve for T we
just divide both sides by 2. So if we divide both sides by 2 we get, T is equal to all of this
business, log base 10 of 7 + 3 all of that over 2, so let me
see if I can remember this, and write it in the actual
box that they gave us. So we want to do log base
10 of 7 and then that + 3 yup it's formatting it
right, divided by 2. So that's what I'm
claiming the T is equal to in terms of Base 10 logarithms. So let me check my answer
and I got it right.