document.write( "Question 381355: How do you solve for X in this equation?: $\"+2ln4x=15+\"$ \n" ); document.write( "

Algebra.Com's Answer #270717 by jsmallt9(3758) $\"\"$ $\"About$

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2ln(4x)=15
\n" ); document.write( "With a variable in the argument of a logarithm, you usually start by transforming the equation into one of the following forms:
\n" ); document.write( "log(expression) = other-expression
\n" ); document.write( "or
\n" ); document.write( "log(expression) = log(other-expression)

\n" ); document.write( "All we have to do to your equation to reach the first form is to divide by 2:
\n" ); document.write( " $\"ln%284x%29+=+15%2F2\"$
\n" ); document.write( "With the first form, the next step is to rewrite the equation in exponential form. In general $\"log%28a%2C+%28p%29%29+=+q\"$ is equivalent to $\"p+=+a%5Eq\"$ . Using this on your equation we get:
\n" ); document.write( " $\"4x+=+e%5E%2815%2F2%29\"$ (Since e is the base of ln.)
\n" ); document.write( "And last of all, we multiply by 1/4. (Dividing by 4 works, too.)
\n" ); document.write( " $\"x+=+%281%2F4%29e%5E%2815%2F2%29\"$
\n" ); document.write( "This is an exact expression for the solution to your equation. \n" ); document.write( "

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