document.write( "Question 1027799: Evaluate using logarithms:\r
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Algebra.Com's Answer #643011 by kev82(151) $\"\"$ $\"About$

You can put this solution on YOUR website!

Have you tried taking the logarithm of both sides? If you don't know what base to use, don't choose one, just use $\"log_b%28x%29\"$ to mean the logarithm of $\"x\"$ to base $\"b\"$ . You will see later that the base you choose is irrelevant.

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We will need a couple of properties of logarithms in order to fully solve this, but the most important one is this:
\n" ); document.write( " $\"+a%2Alog_b%28c%29+=+log_b%28c%5Ea%29+\"$
\n" ); document.write( "If we apply this rule to remove the exponent on the lhs we get $\"+%28a%2B2%29log_b%287%29+\"$ , similarly the rhs is $\"+2a%2A+log_b%289%29+\"$ .

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The resulting equation is actually linear in a, so you should be able to rearrange the equation so that it looks like $\"+a%2AL_1+=+L_2+\"$ where $\"+L_1%2C+L_2+\"$ are expressions in terms of logarithms to base b of numbers.

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You can then get an exact expression for a, $\"+a=L_2%2FL_1+\"$ .

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This expression still has a $\"b\"$ in it, but if you try with different values of $\"b\"$ you should find you always get the same answer.

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See if you can use some of the other properties of logarithms to show that the results will always be the same regardless for what value of $\"b%3E0\"$ that you choose.

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