how

[Guest]

how do you do this problem:

1/16^x+2=32^x-1

 

[stapel]

how do you do this problem

What is to be "done"? (You neglected to include the instructions.)

1/16^x+2=32^x-1

Your formatting is ambiguous. Do you mean any of the following?


. . . . .$\displaystyle \large{\frac{1}{16^x}\,+\,2\,=\,3(2^x)\,-\,1}$


. . . . .$\displaystyle \large{\frac{1}{16^x\,+\,2}\,=\,32^x\,-\,1}$


. . . . .$\displaystyle \large{\frac{1}{16^{x\,+\,2}}\,=\,3(2^{x\,-\,1})}$


Or do you mean something else?

When you reply, please include a clear listing of everything you have tried thus far. Thank you.

Eliz.

 

[soroban]

Hello, bowman1!

I'll take a guess . . .

Could it be . . . \(\displaystyle \L\frac{1}{16^{x+2}} \:= \:32^{x-1}\)

We have: \(\displaystyle \L\,\frac{1}{(2^4)^{x+2}}\;=\;(2^5)^{x-1}\)

Then: \(\displaystyle \L\,\frac{1}{2^{4x+8}}\;=\;2^{5x-5}\;\;\Rightarrow\;\;2^{-4x-8}\;=\;2^{5x-5}\)

Equate exponents: \(\displaystyle \L\,-4x\,-\,8\:=\:5x\,-\,5\;\;\Rightarrow\;\;9x\,=\,-3\;\;\Rightarrow\;\;x\,=\,-\frac{1}{3}\)