Can someone walk me through this exponential question and help check my individual steps?

Question

jojokiw3 · Answer

$$7^{3x-2} = 13^{x+1}$$

jojokiw3 · Answer

Here's my first step: $$\log (7^{3x-2}) = \log(13^{x+1})$$ What's should I try to do next?

jojokiw3 · Answer

$$(3x-2) \log 7 = (x+1) \log 13$$  Is that right?

superdavesuper · Answer

yup - next step is to expand the ( ) on both left n right; then re-group the two terms including x on 1 side n everything else on the other.

jojokiw3 · Answer

Okay how exactly do I do it? Distributive property?

superdavesuper · Answer

correct :)

jojokiw3 · Answer

question, can log 3x become 3 log x?

jojokiw3 · Answer

That technically gets you the same answer right?

superdavesuper · Answer

nope: 3logx=log(x^3)
in any case, plz note that the x is OUTSIDE of the log terms anyway

jojokiw3 · Answer

Here's where I'm stuck, I don't know exactly how or why or how to use distributive here.

superdavesuper · Answer

i will help w the left hand side:

3xlog7 - 2log7 =....

plz do the right hand side

jojokiw3 · Answer

log 7^3x - log 7^2?

superdavesuper · Answer

no need to put the 3x term back into the log function....leave it outside plz

just do the same operation for the right hand side of the =......

jojokiw3 · Answer

xlog13 + log13

superdavesuper · Answer

good :) so there are two terms; one each on the left and right that contains x. move them all on 1 side n the other two terms that do not contain x on the other side...

jojokiw3 · Answer

3xlog 7 - xlog13 = 2log7 + log13

superdavesuper · Answer

VERY good :) now pull the common factor x out on the left side....

jojokiw3 · Answer

$$x = \frac{ 2 \log 7 + \log 13 }{ 3 \log 7 - \log 13 }$$

jojokiw3 · Answer

Right?

superdavesuper · Answer

bingo - u got it :)

jojokiw3 · Answer

Ah thanks for the help. I think I got it now. :)

superdavesuper · Answer

welcome