Solving Exponentiation Problem Without Logarithm

[MatheusMkalo]

Problem in english:
Whereas 2^x = \alpha and 2^y = \beta, write the expression (0,125)^y-2x in function of \alpha and \beta

Without logarithm...

 

[HallsofIvy]

0,125= 1/8= 2^{-3} so (0,125)^{y- 2x}= (2^{-3})^{y-2x}= 2^{-3y+ 6x}. Can you finish it?

 

[MatheusMkalo]

Hehe i found that too but i can't put in function of alfa and beta =/

Answer in book is:
\alpha⁶
\overline{\beta}³

But i don't know how to do.

 

[TylerH]

Hehe i found that too but i can't put in function of alfa and beta =/

Answer in book is:
\alpha⁶
\overline{\beta}³

But i don't know how to do.

Did you know that a^{b-c}=\frac{a^b}{a^c}?

 

[MatheusMkalo]

No i don't see it '-' thank you very much xD