document.write( "Question 689681: please help me solve this problem \r
\n" ); document.write( "\n" ); document.write( "ln (x+5) - ln (x-2) = 2\r
\n" ); document.write( "\n" ); document.write( "x= ?? \n" ); document.write( "
Algebra.Com's Answer #426077 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
ln(x+5) - ln(x-2) = 2
\n" ); document.write( "Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
\n" ); document.write( "log(expression) = number
\n" ); document.write( "or
\n" ); document.write( "log(expression) = log(expression)

\n" ); document.write( "If we can find a way to combine the two logs into one the equation will be in the first form. The two logs are not like terms so we cannot just subtract them. (Like logarithmic terms have the same bases and the same arguments. Your logs have the same base, e, but the arguments, x+5 and x-2, are different.) Fortunately there is a property of logarithms that allow us to combine two logs with a \"-\" between them: \"log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29\". This property requires the same bases and coefficients of 1. Your logs meet both requirements. Using this property we get:
\n" ); document.write( "\"ln%28%28x%2B5%29%2F%28x-2%29%29+=+2\"
\n" ); document.write( "We now have the equation in the first form.

\n" ); document.write( "The next step with this form is to rewrite the equation in exponential form. In general \"log%28a%2C+%28p%29%29+=+q\" is equivalent to \"a%5Eq+=+p\". Using this patter on your equation (and the fact the the base of ln is e):
\n" ); document.write( "\"e%5E2+=+%28x%2B5%29%2F%28x-2%29\"
\n" ); document.write( "Now that the variable is out of the logarithms we can solve for it. We'll start by multiplying each side by x-2 to eliminate the fraction:
\n" ); document.write( "\"e%5E2%28x-2%29+=+x%2B5\"
\n" ); document.write( "\"e%5E2%2Ax-2e%5E2%29+=+x%2B5\"
\n" ); document.write( "Next we'll gather the x terms on one side and the other terms on the other side of the equation. Subtracting x and adding \"2e%5E2\":
\n" ); document.write( "\"e%5E2%2Ax-x%29+=+2e%5E2%2B5\"
\n" ); document.write( "Factoring out x on the left side:
\n" ); document.write( "\"x%28e%5E2-1%29+=+2e%5E2%2B5\"
\n" ); document.write( "And finally divide both sides by \"e%5E2-1\"
\n" ); document.write( "\"x+=+%282e%5E2%2B5%29%2F%28e%5E2-1%29\"
\n" ); document.write( "This is an exact expression for the solution to your equation. If you want/need a decimal approximation then just use your calculator to evaluate the right side. (If your calculator does not have a button for the number \"e\", then use 2.7182818284590452353602874713527 (or some rounded off version of this).) \n" ); document.write( "
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