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:
$\alpha$⁶
$\overline{\beta}$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:
$\alpha$⁶
$\overline{\beta}$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