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

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SUMMARY

The discussion focuses on solving the logarithmic equation log3^(2x-9) - 2log3^x = -2. Participants clarify the manipulation of logarithmic properties, specifically the rule loga^b = b loga, which simplifies the equation to (2x - 9 - 2x) log 3 = -2. This leads to the conclusion that log 3 = 2/9, indicating that the equation cannot be solved for x in a traditional sense. The conversation emphasizes the importance of understanding logarithmic identities and their application in solving equations.

PREREQUISITES
  • Understanding of logarithmic properties, specifically loga^b = b loga.
  • Familiarity with the concept of logarithmic equations.
  • Basic algebra skills for manipulating equations.
  • Knowledge of logarithmic tables for evaluating logarithmic values.
NEXT STEPS
  • Study advanced logarithmic identities and their applications in equations.
  • Learn about the properties of logarithms in different bases.
  • Explore techniques for solving logarithmic equations with multiple variables.
  • Investigate the use of logarithmic tables and calculators for evaluating logarithmic expressions.
USEFUL FOR

Students studying algebra, particularly those focusing on logarithmic functions, educators teaching logarithmic properties, and anyone seeking to enhance their problem-solving skills in mathematics.

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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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