Proving Logarithm Equations: 1/log3a + 1/log4a = 1/log12a

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The discussion centers on proving the logarithmic equation 1/log3a + 1/log4a = 1/log12a. Participants suggest using a common logarithmic base to simplify the left side of the equation, specifically converting to log base 12. One user initially struggles with adding logarithms of different bases but finds clarity after receiving advice to convert all logs to a single base. The conversation highlights the importance of understanding logarithmic properties and the utility of consistent bases in solving such equations. Ultimately, the problem is resolved through the suggested method, demonstrating the simplicity of the solution.
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Homework Statement



Prove:

1/log3a + 1/log4a = 1/log12a

Homework Equations



ay=x
Logarithms rules (addition, subtraction, power, etc.)

logax=logbx/logba

The Attempt at a Solution

Left Side:

1/log3a + 1/log4a
=log3a+log4a/log12a (via common denominator)

The problem is how to add logarithms with different bases. I tried converting the log3a to log4a (I get log4a/log43). After that, I subbed it back into the equation.

=log4a/log43+log4a

But I don't think that gets me anywhere...

Right side still remains the same (1/log12a)

Any help is appreciated! Thanks in advance :).

P.S. What program do people use to make their equations look so neat (the fraction looks real - ex. 1/4 really looks like 1 (horizontal line) 4)?
 
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Let's just convert everything to a single log base, like log_12. E.g. log_3(a)=log_12(a)/log_12(3).
 
Oh, wow. Sigh, I hate it when you take the wrong approach in proving Left Side equals Right Side.

Thanks for the help! Can't believe it was so simple after your suggestion :).
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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