Solving Exponentiation Problem Without Logarithm

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Homework Help Overview

The discussion revolves around an exponentiation problem involving the expressions defined by the variables α and β, specifically focusing on rewriting the expression (0.125)^(y-2x) in terms of these variables without using logarithms.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the transformation of the expression using properties of exponents, with one participant noting the equivalence of 0.125 to 2^{-3}. There is a question about how to express the result in terms of α and β, with references to a book answer that includes these variables.

Discussion Status

The discussion is ongoing, with participants sharing their findings and expressing uncertainty about how to proceed further. Some guidance has been offered regarding exponent properties, but no consensus or complete solution has been reached.

Contextual Notes

Participants are working under the constraint of not using logarithms, which may limit their approaches to the problem. There is also a reference to a specific answer from a book, indicating a potential expectation for a particular form of the solution.

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

Without logarithm...
 

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[itex]0,125= 1/8= 2^{-3}[/itex] so [itex](0,125)^{y- 2x}= (2^{-3})^{y-2x}= 2^{-3y+ 6x}[/itex]. Can you finish it?
 
Hehe i found that too but i can't put in function of alfa and beta =/

Answer in book is:
[itex]\alpha[/itex]6
[itex]\overline{\beta}[/itex]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:
[itex]\alpha[/itex]6
[itex]\overline{\beta}[/itex]3

But i don't know how to do.

Did you know that [itex]a^{b-c}=\frac{a^b}{a^c}[/itex]?
 
No i don't see it '-' thank you very much xD
 
Last edited:

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