document.write( "Question 359742: Solve for x:\"%28logx%29%5E3=logx%5E4\" \n" ); document.write( "
Algebra.Com's Answer #256902 by jsmallt9(3758)\"\" \"About 
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\"%28log%28%28x%29%29%29%5E3=log%28%28x%5E4%29%29\"
\n" ); document.write( "This equation can be solved if we rewrite it in terms of log(x). The right side of the equation can be rewritten in terms of log(x) if we use a property of logarithms, \"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\", to move the exponent of \"x%5E4\" out in front:
\n" ); document.write( "\"%28log%28%28x%29%29%29%5E3=4%2Alog%28%28x%29%29\"
\n" ); document.write( "If you can't yet see how to solve this, perhaps a temporary variable will help. Let z = log(x). Substituting z in for log(x) we get:
\n" ); document.write( "\"z%5E3+=+4z\"
\n" ); document.write( "To solve a 3rd degree equation like this, we want one side to be zero. So we'll subtract 4z from each side:
\n" ); document.write( "\"z%5E3+-+4z+=+0\"
\n" ); document.write( "And then we factor. First the Greatest Common Factor (which is z):
\n" ); document.write( "\"z%28z%5E2+-+4%29+=+0\"
\n" ); document.write( "Then we can use the difference of squares pattern, \"a%5E2+-+b%5E2+=+%28a%2Bb%29%28a-b%29\", to factor \"z%5E2+-4\" (with \"a\" being \"z\" and \"b\" being 2):
\n" ); document.write( "\"z%28z%2B2%29%28z-2%29+=+0\"
\n" ); document.write( "From the Zero Product Property we know that this (or any) product can be zero only if one of the factors is zero. So
\n" ); document.write( "z = 0 or z+2 = 0 or z-2 = 0
\n" ); document.write( "Solving these we get:
\n" ); document.write( "z = 0 or z = -2 or z = 2
\n" ); document.write( "Of course we do not care what z is. We want to know what x is. So at this point we substitute back log(x) for z:
\n" ); document.write( "log(x) = 0 or log(x) = -2 or log(x) = 2
\n" ); document.write( "If you understand what logarithms are then you may be able to figure out what x is for each of these. If not, then the next step is to rewrite these in exponential form. In general, \"log%28a%2C+%28p%29%29+=+q\" can be rewritten as \"a%5Eq+=+p\". So for your equations, with base 10 logarithms, we get:
\n" ); document.write( "\"10%5E0+=+x\" or \"10%5E%28-2%29+=+x\" or \"10%5E2+=+x\"
\n" ); document.write( "Simplifying these powers of 10 we get:
\n" ); document.write( "1 = x or 1/100 = x or 100 = x
\n" ); document.write( "So there are three solutions to your equation.

\n" ); document.write( "P.S. With some practice on equations like this, you will be able to work without the temporary variable. You will see how to go from
\n" ); document.write( "\"%28log%28%28x%29%29%29%5E3=4%2Alog%28%28x%29%29\"
\n" ); document.write( "to
\n" ); document.write( "\"%28log%28%28x%29%29%29%5E3+-+4%2Alog%28%28x%29%29+=+0\"
\n" ); document.write( "to
\n" ); document.write( "\"log%28%28x%29%29%2A%28%28log%28%28x%29%29%29%5E2+-+4%29+=+0\"
\n" ); document.write( "etc. \n" ); document.write( "
\n" );