How Do You Solve Logarithmic Equations with Coefficients?

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Homework Help Overview

The discussion revolves around solving a logarithmic equation involving coefficients, specifically the equation log6(2x-5)+1=log6(7x+10). Participants are exploring methods to manipulate the equation to isolate the variable.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants suggest exponentiating both sides to eliminate the logarithm, while others express confusion about handling the additional constant on the left side of the equation. There is also a hint provided regarding the properties of logarithms that may assist in simplifying the equation.

Discussion Status

The discussion includes attempts to solve the equation, with one participant successfully arriving at a solution after applying a method involving logarithmic properties. However, there remains a focus on understanding the manipulation of the equation, particularly regarding the constant term.

Contextual Notes

There are no explicit constraints mentioned, but participants are navigating the complexities of logarithmic properties and their application in solving the equation.

darshanpatel
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Homework Statement



Solve:

log6(2x-5)+1=log6(7x+10)

Homework Equations



-None-

The Attempt at a Solution



I know we will need to exponentiate both side by 6 to get rid of the log, or either take another log.

I need help with getting rid of the +1 on the left side of the = sign because that is what is baffling me.
 
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Hint: [itex]6^{1+a} = 6 \times 6^a[/itex], doesn't it?
 
log6(2x-5)+1=log6(7x+10)

6log6(2x-5)+1=6log6(7x+10)

6(2x-5)=7x+10

12x-30=7x+10

5x=40

x=8

I checked it and it worked, thanks for the help man!
 
darshanpatel said:

Homework Statement



Solve:

log6(2x-5)+1=log6(7x+10)

Homework Equations



-None-

The Attempt at a Solution



I know we will need to exponentiate both side by 6 to get rid of the log, or either take another log.


I need help with getting rid of the +1 on the left side of the = sign because that is what is baffling me.

log66 = ?

Chet
 

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