document.write( "Question 258161: Find the exact solution(s) to the equation $\"e%5E%282x%29-2e%5E%28x%29-35=0\"$ \n" ); document.write( "

Algebra.Com's Answer #190032 by drk(1908) $\"\"$ $\"About$

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The original equation was
\n" ); document.write( "(i) $\"e%5E%282x%29-2e%5E%28x%29-35=0\"$
\n" ); document.write( "Let Y = e^x. then Y^2 = e^(2x). We get
\n" ); document.write( "(ii) $\"Y%5E2-2Y-35=0\"$
\n" ); document.write( "factoring, we get
\n" ); document.write( "(iii) $\"%28Y-7%29%28Y%2B5%29+=+0\"$
\n" ); document.write( "solving
\n" ); document.write( "Y = 7
\n" ); document.write( "OR
\n" ); document.write( "Y = -5
\n" ); document.write( "substituting e^x for Y, we get
\n" ); document.write( "(iv) $\"e%5Ex+=+7\"$
\n" ); document.write( "so, x = ln(7)
\n" ); document.write( "--
\n" ); document.write( "(v) $\"e%5Ex+=+-5\"$
\n" ); document.write( "gives no solution. so, we are left with
\n" ); document.write( "x = ln(7)
\n" ); document.write( " \n" ); document.write( "

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