Solve Log EQ: log3^(2x-9)-2xlog3=-2

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Homework Help Overview

The discussion revolves around solving a logarithmic equation involving the logarithm base 3, specifically the equation log3^(2x-9) - 2log3^x = -2. Participants are exploring the manipulation of logarithmic expressions and the validity of their steps in the context of logarithmic properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of moving terms in logarithmic expressions, questioning whether the manipulation of 2log3^x to 2xlog3 is appropriate. There are attempts to simplify the equation and clarify the steps taken, including the application of logarithmic rules.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning and clarifying the original equation. Some participants suggest that the equation may not be solvable for x, while others explore different interpretations of the logarithmic properties involved.

Contextual Notes

There is an ongoing examination of the assumptions related to the properties of logarithms and the implications of the equation's structure. Participants are also reflecting on the potential for multiple interpretations of the logarithmic expressions involved.

juliany
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Homework Statement


Solve: log3^(2x-9)-2log3^x=-2


Homework Equations


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The Attempt at a Solution


I am confused with one part of this equation.
With the 2log3^x, can you move the x to the front to make it 2x log3?
 
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Yes, that's valid.
 
So that would make it 2x-9log3-2xlog3=-2
Therefore: 2x-9-2x=-2?
 
where did the log 3 go in your last step?
 
:blushing:Woops, was thinking the same base rule.

Can you go: log3^(2x-9)-log3^x^2=-2
Therefore: Log3 (2x-9)/x^2=-2?
 
2 \log 3^{x} is \log 3^{2x}. Just to make sure, was this the original equation?

\log3^{2x-9}-2\log 3^x=-2​

Because it cannot be solved for x.
 
Last edited:
Yes it was, how did you figure out that you can't solve for x?
 
Using the rule \log a^{b} = b \log a, the equation becomes (2x - 9 - 2x) \log 3 = - 2, and the 2x and -2x cancel out. Then you get the equation \log 3 = 2/9. This is true for an appropriate base of the logarithm, which you can find using a log table.
 
:smile:Ok, thanks a lot.
 

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