document.write( "Question 1087097: I'm trying to figure out how to get the E alone in this equation. R=(2/3)log E-3 \n" ); document.write( "
Algebra.Com's Answer #701381 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
IF you have this:\r
\n" ); document.write( "\n" ); document.write( "\"R=%282%2F3%29log%28+E%29-3+\"\r
\n" ); document.write( "\n" ); document.write( "\"R%2B3=%282%2F3%29log%28+E%29+\"\r
\n" ); document.write( "\n" ); document.write( "\"%28R%2B3%29%2F%282%2F3%29=log%28+E%29+\"\r
\n" ); document.write( "\n" ); document.write( "\"3%28R%2B3%29%2F2=log%28+E%29+\"\r
\n" ); document.write( "\n" ); document.write( "\"log%28+E%29=+3%28R%2B3%29%2F2\".....since \"y+=+log%28b%2Cx%29\" is equivalent to \"x+=+b%5Ey\", and in your case \"y+=%283R%2B9%29%2F2\", base \"b=10\", and \"x=E\"\r
\n" ); document.write( "\n" ); document.write( "so, you will have:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"highlight%28E=10%5E%283%28R%2B3%29%2F2%29%29\"\r
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