document.write( "Question 941697: what is the inverse of y= log2 (8^(x-1)) \n" ); document.write( "
Algebra.Com's Answer #574064 by MathLover1(20850)\"\" \"About 
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to find the inverse of \"y=++log%282%2C8%5E%28x-1%29%29\", first swap \"x\" and \"y\"\r
\n" ); document.write( "\n" ); document.write( "\"x=++log%282%2C8%5E%28y-1%29%29\"...now solve for \"y\"\r
\n" ); document.write( "\n" ); document.write( "\"x=++log%282%2C8%5E%28y-1%29%29\"...change to base \"10\"\r
\n" ); document.write( "\n" ); document.write( "\"x=++log%288%5E%28y-1%29%29%2Flog%282%29\"\r
\n" ); document.write( "\n" ); document.write( "\"x%2Alog%282%29=++log%288%5E%28y-1%29%29\"\r
\n" ); document.write( "\n" ); document.write( "\"log%282%5Ex%29=++log%288%5E%28y-1%29%29\"...since log same, we have\r
\n" ); document.write( "\n" ); document.write( "\"2%5Ex=+8%5E%28y-1%29\"....we can write \"8\" as \"2%5E3\"\r
\n" ); document.write( "\n" ); document.write( "\"2%5Ex=+%282%5E3%29%5E%28y-1%29\"\r
\n" ); document.write( "\n" ); document.write( "\"2%5Ex=+2%5E%283%28y-1%29%29\".....if bases same then exponents are same too\r
\n" ); document.write( "\n" ); document.write( "\"x=+3%28y-1%29\"\r
\n" ); document.write( "\n" ); document.write( "\"x=+3y-3\"\r
\n" ); document.write( "\n" ); document.write( "\"x%2B3=+3y\"\r
\n" ); document.write( "\n" ); document.write( "\"y=x%2F3%2B1\"..=>the inverse of \"y=++log%282%2C8%5E%28x-1%29%29\"\r
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