I have been doing some review questions for a course and can't find anything on how to do this, just remember that I need to change it into a log question somehow:

3(5^x)-1= 2

5^(x)+2

Would appreciate some hints as to how to solve it.

Thanks

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- Mar 6th 2010, 10:21 PM #1
## Log question

I have been doing some review questions for a course and can't find anything on how to do this, just remember that I need to change it into a log question somehow:

__3(5^x)-1__= 2

5^(x)+2

Would appreciate some hints as to how to solve it.

Thanks

- Mar 6th 2010, 10:27 PM #2
I take it that this is

$\displaystyle \frac{3\left(5^x\right) - 1}{5^x + 2} = 2$

$\displaystyle 3\left(5^x\right) - 1 = 2\left(5^x + 2\right)$

$\displaystyle 3\left(5^x\right) - 1 = 2\left(5^x\right) + 4$

$\displaystyle 5^x = 5$.

What do you think $\displaystyle x$ has to be?

- Mar 7th 2010, 01:10 AM #3