document.write( "Question 404441: solve for x to the nearest thousandth: ln(5x+20)=2 \n" ); document.write( "

Algebra.Com's Answer #285828 by jsmallt9(3758) $\"\"$ $\"About$

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ln(5x+20)=2
\n" ); document.write( "Solving equations where the variable is in the argument of a logarithm, like this equation, usually starts with 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( "Your equation is already in the first form. So we can proceed to the next step, which 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 pattern on your equation (and using the fact that the base of ln is \"e\") we get:
\n" ); document.write( " $\"5x%2B20+=+e%5E2\"$
\n" ); document.write( "We can now solve for x. Subtracting 20 from each side we get:
\n" ); document.write( " $\"5x+=+e%5E2+-+20\"$
\n" ); document.write( "Dividing by 5 we get:
\n" ); document.write( " $\"x+=+%28e%5E2+-+20%29%2F5\"$
\n" ); document.write( "This is an exact expression of the solution to your equation. You are asked to find a decimal (approximation). So get out your calculator. (If your calculator does not have a button for the number \"e\", use 2.7182818284590451 (or some rounded off version of it).) \n" ); document.write( "

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