document.write( "Question 534035: I am stuck on this problem. I am supposed to solve for x:\r
\n" ); document.write( "\n" ); document.write( "e^0.5x=3^(1-2x)\r
\n" ); document.write( "\n" ); document.write( "Thanks for any assistance you can give. \n" ); document.write( "

Algebra.Com's Answer #351246 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
Solve for x:
\n" ); document.write( " $\"e%5E0.5x+=+3%5E%281-2x%29\"$ Take the natural log of both sides.
\n" ); document.write( " $\"Ln%28e%5E%280.5x%29%29+=+Ln%283%5E%281-2x%29%29\"$ Apply the \"power rule\" for logarithms.
\n" ); document.write( " $\"0.5xLn%28e%29+=+%281-2x%29Ln%283%29\"$ Substitute $\"Ln%28e%29+=+1\"$ and $\"Ln%283%29+=+1.1\"$ approx.
\n" ); document.write( " $\"0.5x+=+%281-2x%29%281.1%29\"$ Evaluate.
\n" ); document.write( " $\"0.5x+=+1.1-2.2x\"$ Add 2.2x to both sides.
\n" ); document.write( " $\"2.7x+=+1.1\"$ Divide both sides by 2.7
\n" ); document.write( " $\"x+=+1.1%2F2.7\"$
\n" ); document.write( " $\"x+=+11%2F27\"$
\n" ); document.write( " $\"x+=+0.407\"$ approx. \n" ); document.write( "

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