Solve log_b(xy): Express in terms of p, q

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Homework Help Overview

The discussion revolves around expressing the logarithm of the product of two variables, log_b(xy), in terms of given logarithmic expressions involving x and y. The subject area includes logarithmic properties and algebraic manipulation.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial expressions involving log_b(x^3y) and log_b(y/x^2) and explore how to derive log_b(xy) from these. Some participants suggest finding log x and log y in terms of p and q as a preliminary step. Others express confusion about the process and seek clarification on how to proceed.

Discussion Status

The discussion is ongoing, with participants sharing different approaches and attempting to clarify their understanding. Some have offered partial insights, while others are still seeking guidance on the next steps. There is no explicit consensus yet on the best method to express log_b(xy).

Contextual Notes

Participants are working under the constraints of the given logarithmic equations and are attempting to manipulate these without additional information or solutions provided. The nature of the problem suggests a focus on algebraic relationships between the variables involved.

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[tex]\text{Given that } log_b(x^3y)=p \text{ and } log_b(\frac{y}{x^2}) = q \text{ express } log_b(xy) \text{ in terms of p and q.}[/tex]
[tex]\text{ This is what I've done. }[/tex]
[tex]log_b(x^3y)=p----(1)[/tex]
[tex]log_b(\frac{y}{x^2})=q ----(2)[/tex]
1 - 2
[tex]5log_bx=p-q[/tex]
[tex]\text{ I'm stuck here. What do I do next? }[/tex]
 
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Find log x and log y in terms of p and q first
then split log(xy)
 
primarygun said:
Find log x and log y in terms of p and q first
How do I do that?Could you show me the steps?
 
footprints said:
How do I do that?Could you show me the steps?

[tex]b^{p}=x^{3}y (1);b^{q}=xy^{-2} (2)[/tex]
(1)=>[tex]b^{2p}=x^{6}y^{2} (3)[/tex]
(2)*(3)=> [tex]x^{7}=b^{2p+q} (4)[/tex] => log_{b}x=...

Did u see a pattern??
 
Nope. I don't see a pattern. But I figured how to do it. Thanks for help.
 

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