How to solve for x?
11.04=10(1.02)^x

Question

How to solve for x?
11.04=10(1.02)^x

phi · Answer

I would first divide both sides by 10.

anonymous · Answer

So, then I have 1.104= (1.02)^x right?

phi · Answer

yes.  now "take the log" of both sides:
log(stuff on left)  = log(stuff on right)

phi · Answer

we can assume log base 10

on the right side you can use this "rule":
$$ \log(a^b)= b\log(a) $$

phi · Answer

$$ \log\left(1.104 \right)= \log\left(1.02^x \right) $$

anonymous · Answer

That's kind of over my head... I'm not sure what the "log" is?

phi · Answer

You need to use logarithms to solve it. You must have studied this. If you need a refresher, see https://www.khanacademy.org/math/algebra2/logarithms-tutorial/logarithm_basics/v/logarithms

phi · Answer

and here is an example of solving this type of equation https://www.khanacademy.org/math/algebra2/logarithms-tutorial/logarithm_basics/v/exponential-equation though in your problem, you will need a calculator.

anonymous · Answer

Okay so at this point, I have to solve log1.02 1.104=x ?

anonymous · Answer

Which is 1.02^5?

phi · Answer

how did you get that ?

anonymous · Answer

It's what they did in that first khan academy video, and I just substituted my numbers in...

phi · Answer

$$ \log\left(1.104 \right)= \log\left(1.02^x \right) $$
re-write the right side using $ \log(a^b) = b \log(a) $
can you do that ?

phi · Answer

here is an example with just numbers:
log(10) = 1  (try it with a calculator)
log(10^2) = 2 log(10) = 2*1 = 2
as a test, we know 10^2 = 10*10 = 100 and log(100)= 2
the rule works!

phi · Answer

in your problem match the "a" with 1.02  and "b" with x

anonymous · Answer

I'm sorry I'm not quite getting what you're telling me, but that khan academy link you sent really helped. I got the answer right... Thanks for helping!