document.write( "Question 27407: I was wondering if you could perhapse teach me how to solve this question step by step. thank you.\r
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\n" ); document.write( "\n" ); document.write( "10^(x-3)=e^(2x+5) \n" ); document.write( "
Algebra.Com's Answer #14891 by sdmmadam@yahoo.com(530)\"\" \"About 
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\n" ); document.write( "10^(x-3)=e^(2x+5)\r
\n" ); document.write( "\n" ); document.write( "Given 10^(x-3)=e^(2x+5)
\n" ); document.write( "Taking logarithm on both the sides to the base e (the Naperian base)
\n" ); document.write( "loge[10^(x-3)] = loge[e^(2x+5)]
\n" ); document.write( "which implies (x-3)log10 = (2x+5)loge (the base being e on both the sides)
\n" ); document.write( "Here we have taken logarithm on both the sides with an idea to bring the power down using the formula loge(m^n) = n X loge(m)
\n" ); document.write( "Therefore (x-3)X loge(10) = (2x+5)X loge(e)
\n" ); document.write( " (x-3)y = (2x+5)X 1 (log e to the same base e is 1)
\n" ); document.write( "We have put loge(10) = y for convenience.
\n" ); document.write( "So, we have (x-3)y = (2x+5)
\n" ); document.write( " which is xy -3y = 2x +5
\n" ); document.write( " xy - 2x = 5 + 3y (transposing, change side then change sign)
\n" ); document.write( " x(y-2) = 5 + 3y
\n" ); document.write( "Dividing by (y-2)(which is not zero)
\n" ); document.write( " x = (5+3y)/(y-2) where y = loge(10)
\n" ); document.write( "Note: (y-2) is not zero because y = loge(10) is not equal to 2
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