document.write( "Question 613779: 8e^2x=136 \n" ); document.write( "

Algebra.Com's Answer #386331 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"8e%5E%282x%29=136\"$
\n" ); document.write( "First, isolate the base and its exponent by dividing both sides by 8:
\n" ); document.write( " $\"e%5E%282x%29=17\"$
\n" ); document.write( "Now find the log of each side. While any base of logarithm can be used, it makes the most sense to use base e logs (aka ln) because

matching the base of the logarithm to the base of the exponent will lead to a simpler answer
your calculator \"knows\" base e logs. This will make it easy to find a decimal approximation of the answer if desired.

\n" ); document.write( " $\"ln%28e%5E%282x%29%29+=+ln%2817%29\"$

\n" ); document.write( "Next use a property of logarithms, $\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"$ , which allows us to move the exponent of an argument out in front of the logarithm. (It is this property that is the very reason we use logarithms on these equations. Being able to move the exponent, where the variable is, like this the variable is no longer in an exponent. We can then use \"regular\" algebra to solve for the variable.) Using this property on our equation we get:
\n" ); document.write( " $\"2x%2Aln%28e%29+=+ln%2817%29\"$

\n" ); document.write( "Since $\"log%28a%2C+%28a%29%29+=+1\"$ for all bases, the ln(e) = 1. (This is why matching the base of the logarithm to the base of the exponent is advantageous.) The left side becomes:
\n" ); document.write( " $\"2x+=+ln%2817%29\"$
\n" ); document.write( "Now we just divide by 2:
\n" ); document.write( " $\"x+=+ln%2817%29%2F2\"$
\n" ); document.write( "This is an exact expression for the solution to this problem. If you want/need a decimal approximation you can use your calculator to find ln(17) and then to divide by 2. \n" ); document.write( "

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