document.write( "Question 779140: If c^2=a^2-b^2, show that 1/log c (to base a+b)+1/log c (to base a-b)=2 \n" ); document.write( "

Algebra.Com's Answer #475168 by jsmallt9(3758) $\"\"$ $\"About$

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The keys to this problems are:

Recognizing that the right side of $\"c%5E2=a%5E2-b%5E2\"$ is a difference of squares. And its factors, (a+b) and (a-b), are the bases of the logs in the final equation.
The denominators in $\"1%2Flog%28a%2Bb%2C+%28c%29%29+%2B+1%2Flog%28a-b%2C+%28c%29%29\"$ look like they may have come from the change of base formula.
If the denominators are from the change of base then the numerators should be logs of the same base. We can rewrite the numerators, the 1's, as logarithms:
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Looking at these fractions this way we can see that they come from converting base c logs.

So our plan is to factor the difference of squares, find base c logs of both sides of $\"c%5E2=%28a%2Bb%29%28a-b%29\"$ and then convert the logs:
\n" ); document.write( " $\"c%5E2=a%5E2-b%5E2\"$
\n" ); document.write( " $\"c%5E2=%28a%2Bb%29%28a-b%29\"$
\n" ); document.write( " $\"log%28c%2C+%28c%5E2%29%29=log%28c%2C%28%28a%2Bb%29%28a-b%29%29%29\"$
\n" ); document.write( "Using properties of logarithms we can rewrite each of the equation:
\n" ); document.write( " $\"2%2Alog%28c%2C+%28c%29%29+=+log%28c%2C+%28a%2Bb%29%29+%2B+log%28c%2C+%28a-b%29%29\"$
\n" ); document.write( "The log on the left side is 1:
\n" ); document.write( " $\"2+=+log%28c%2C+%28a%2Bb%29%29+%2B+log%28c%2C+%28a-b%29%29\"$
\n" ); document.write( "Using the base conversion formula we will convert the first log on the right from base c to base a+b and the second log from base c to base a-b:
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\n" ); document.write( "The numerators on the right are 1's:
\n" ); document.write( " $\"2+=+1%2Flog%28a%2Bb%2C+%28c%29%29+%2B+1%2Flog%28a-b%2C+%28c%29%29\"$
\n" ); document.write( "And we are finished! \n" ); document.write( "

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