document.write( "Question 71846: Solve for x in the following eguation.\r
\n" ); document.write( "\n" ); document.write( "5=9^x-1\r
\n" ); document.write( "\n" ); document.write( "(as in x-1 are exponents.)
\n" ); document.write( "I can't figure out where the log goes... :(
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Algebra.Com's Answer #51380 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
The log of base 9 goes on both sides. The log undoes an exponent, so we use the log on the entire right side, but we must do it to both sides.
\n" ); document.write( "\"5=9%5E%28x-1%29\"
\n" ); document.write( "\"log_%5B9%5D%285%29=log_%5B9%5D%289%5E%28x-1%29%29\"Take the log of both sides.The log simply cancels out the base 9
\n" ); document.write( "\"log_%5B9%5D%285%29=+x-1\"Add 1 to both sides
\n" ); document.write( "\"log_%5B9%5D%285%29%2B1=x\"Since log_[9](5) is a number we can find x
\n" ); document.write( "Use the change of base formula
\n" ); document.write( "\"log_%5Bb%5D%28x%29=log%28x%29%2Flog%28b%29\"Where b is the base and x is the argument. Remember the base of the log is assumed to be 10 if not specified.
\n" ); document.write( "\"x=log%285%29%2Flog%289%29%2B1\"So this can be evaluated on a calculator
\n" ); document.write( "\"x=0.69897%2F0.95424+%2B+1\"
\n" ); document.write( "\"x=0.73248+%2B+1\"
\n" ); document.write( "\"x=1.73248\"
\n" ); document.write( "Check:
\n" ); document.write( "\"5=9%5E%281.73248-1%29\"
\n" ); document.write( "\"5=9%5E%280.73248%29\"
\n" ); document.write( "\"5=4.99993\" Which is close enough to 5 (there are some roundoff errors).
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