SOLUTION: I am trying to solve (1/8)^(2x-3) = 16^(x+1)

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Question 983070: I am trying to solve (1/8)^(2x-3) = 16^(x+1)
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!
I won't do your problem for you.  Instead here is one EXACTLY like yours.
Use it as a model to solve yours:



Write 27 as 3³ and 9 as (3²)



Write  as 



remove the parentheses by multiplying the exponents:



Since the bases are the same, and they are positive
and not equal to 1, we may equate the exponents only









Edwin
Answer by ikleyn(52802)   (Show Source): You can put this solution on YOUR website!

Is this the equation you need to solve?

= .

If so,  then let us write it in terms of degrees of  2:

= .

It gives

-3*(2x-3) = 4*(x+1).

To solve the last equation,  simplify it step by step:

-6x + 9 = 4x + 4,

-6x - 4x = 4 - 9,

-10x = -5,

x = = .

Check.  The left side of the original equation at  x =   is   = = = .

            The right side of the original equation at  x =   is   = = 64.

Answer.  x = .


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