1. ## Solving Exponential Equations

Hi I apologize for yet another question but my lesson book is terribly vague and does not provide examples so I'm not quite sure how to proceed with this problem.

Problem:

$(2^{2x+1})(4^{3x+3})=128^x$

I haven't gotten into logs yet so I obviously need like bases so I've done this:

$2^{(2x+1)}2^{2(3x+3)}=2^{7(x)}$

Which gives me

$(2x+1)(6x+6)=7x$

if what I did is right than I'm thinking I should expand the brackets move the 7x over to make it =0 and factor?

Thanks for taking a look!

2. Yep, that's it

3. Originally Posted by Substince
Hi I apologize for yet another question but my lesson book is terribly vague and does not provide examples so I'm not quite sure how to proceed with this problem.

Problem:

$2^{2x+1})(4^{3x+3})=128^x$

I haven't gotten into logs yet so I obviously need like bases so I've done this:

$2^{(2x4)}2^{2(3x+3)}=2^{7(x)}$

Which gives me

$(2x+1)(6x+6)=7x$

if what I did is right than I'm thinking I should expand the brackets move the 7x over to make it =0 and factor?

Thanks for taking a look!
re: the left side of your equation. note that ...

$2^{2x+1} \cdot 2^{6x+6} = 2^{(2x+1)+(6x+6)}$