document.write( "Question 246776: How do I figure out this problem?
\n" ); document.write( "Log subscript x, 4=1/3 \n" ); document.write( "

Algebra.Com's Answer #180111 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"log%28x%2C+%284%29%29+=+1%2F3\"$
\n" ); document.write( "When solving for a variable which is in the base or the argument to a logarithm, you will, as a general rule, rewrite the equation in exponential form. To do this we need to remember that $\"log%28a%2C+%28p%29%29+=+q\"$ is equivalent to $\"p+=+a%5Eq\"$ . Using this on your equation we get:
\n" ); document.write( " $\"4+=+x%5E%281%2F3%29\"$
\n" ); document.write( "Our answer will be \"x = some-number\" or \"some-number = x\". These x's do not have a visible exponent. But they do have an exponent. Invisible exponents are 1's. So we want to change $\"x%5E%281%2F3%29\"$ to $\"x%5E1\"$ somehow. This can be done easily if we know how exponents work. (Hint: $\"%28a%5Ep%29%5Eq+=+a%5E%28p%2Aq%29\"$ ) We just raise each side to the 3rd power:
\n" ); document.write( " $\"%284%29%5E3+=+%28x%5E%281%2F3%29%29%5E3\"$
\n" ); document.write( "which gives us:
\n" ); document.write( " $\"64+=+x%5E1\"$
\n" ); document.write( "or
\n" ); document.write( " $\"64+=+x\"$
\n" ); document.write( " \n" ); document.write( "

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