document.write( "Question 205449This question is from textbook Functions modeling Change
\n" ); document.write( ": I'm having trouble with sovling the log equations. i have two questions. 3^(1-2x)=e^(0.5x). and log(x+1)+ log(x-1)=log(3) i know that you start off by taking the natural log of both sides but after that i'm stuck. i don't know which logarithm property to you. your help is much appreciated \n" ); document.write( "

Algebra.Com's Answer #155117 by Alan3354(69443) $\"\"$ $\"About$

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3^(1-2x)=e^(0.5x)
\n" ); document.write( "ln(3)*(1-2x) = 0.5x
\n" ); document.write( "ln(3) - 2x*ln(3) = 0.5x
\n" ); document.write( "ln(3) = 0.5x + 2x*ln(3)
\n" ); document.write( "ln(3) = x*(0.5 + 2ln(3))
\n" ); document.write( "x = ln(3)/(0.5 + 2ln(3))
\n" ); document.write( "Everything on the right is a constant.
\n" ); document.write( "--------------
\n" ); document.write( "log(x+1)+ log(x-1)=log(3)
\n" ); document.write( "log((x+1)*(x-1)) = log(3)
\n" ); document.write( "(x+1)*(x-1) = 3
\n" ); document.write( "x^2 - 1 = 3
\n" ); document.write( "x^2 = 4
\n" ); document.write( "x = +2
\n" ); document.write( "x = -2
\n" ); document.write( "The -2 won't work, as it gives log(-3), so
\n" ); document.write( "x = 2
\n" ); document.write( " \n" ); document.write( "

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