How to Solve Logarithmic Equations with Different Bases?

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Homework Help Overview

The discussion revolves around solving the equation log2x + log4x = 5, focusing on logarithmic equations with different bases. Participants express uncertainty regarding the approach to take, particularly when dealing with varying bases in logarithmic functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest converting logarithms to the same base to facilitate solving the equation. Others question the interpretation of the equation, considering different ways to express the logarithmic terms. There is also a mention of using properties of logarithms to simplify the problem.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations and approaches. Some guidance has been offered regarding changing the base of logarithms, but there is no explicit consensus on the best method to proceed.

Contextual Notes

There appears to be confusion regarding the notation used in the logarithmic equation, which may affect the interpretation and subsequent approach to solving it. Additionally, some participants are seeking clarification on how to properly present their questions in the forum.

emma3001
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Solve the equation log2x + log4x= 5. To start, should I change this to an exponential... I am stuck because I have only done log questions that have the same base.
 
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Convert them to the same base log_a(x)=log_b(x)/log_b(a).
 
emma3001 said:
Solve the equation log2x + log4x= 5. To start, should I change this to an exponential... I am stuck because I have only done log questions that have the same base.
What do you mean by "log2x+ log4x= 5"? I would interpret that as log(2x)+ log(4x)= 5 so log(6x)= 5 which is easy. If you mean "log_2(x)+ log_4(x)= 5" then use Dick's hint.
log_4(x)= log_2(x)/log_2(4)= ?
 
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Dick said:
Convert them to the same base log_a(x)=log_b(x)/log_b(a).

can u help solving a problem?
 
can u help solve a problem?
 
Can u post the problem? Under Forum Tools, "Post a New Thread". That's your first step.
 

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