Solving Exponentiation Problem Without Logarithm

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To solve the expression (0.125)^(y-2x) in terms of α and β, first recognize that 0.125 equals 2^(-3). This allows the expression to be rewritten as (2^(-3))^(y-2x), simplifying to 2^(-3y + 6x). The challenge is to express this in terms of α (2x) and β (2y). The final answer is α^6 / β^3, which can be derived using the property a^(b-c) = a^b / a^c. Understanding this property is crucial for completing the solution.
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Problem in english:
Whereas 2x = \alpha and 2y = \beta, write the expression (0,125)y-2x in function of \alpha and \beta

Without logarithm...
 

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0,125= 1/8= 2^{-3} so (0,125)^{y- 2x}= (2^{-3})^{y-2x}= 2^{-3y+ 6x}. Can you finish it?
 
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.
 
Last edited:
MatheusMkalo said:
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}?
 
No i don't see it '-' thank you very much xD
 
Last edited:

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