document.write( "Question 354710: solve for x.... logx64=-6 \n" ); document.write( "

Algebra.Com's Answer #253765 by jsmallt9(3758) $\"\"$ $\"About$

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With an equation of the form
\n" ); document.write( "log(expression) = other-expression
\n" ); document.write( "and the variable in the argument or base of the logarithm, then you will usually rewrite the equation in exponential form. In general, $\"log%28a%2C+%28p%29%29+=+q\"$ is equivalent to $\"a%5Eq+=+p\"$ . Using this on your equation we get:
\n" ); document.write( " $\"x%5E%28-6%29+=+64\"$
\n" ); document.write( "Rewriting this with a positive exponent we get:
\n" ); document.write( " $\"1%2Fx%5E6+=+64\"$
\n" ); document.write( "We can eliminate the fraction by multiplying both sides by $\"x%5E6\"$ :
\n" ); document.write( " $\"1+=+64x%5E6\"$
\n" ); document.write( "Next we can divide both sides by 64:
\n" ); document.write( " $\"1%2F64+=+x%5E6\"$
\n" ); document.write( "To find x we will find the 6th root of each side:
\n" ); document.write( " $\"root%286%2C+1%2F64%29+=+root%286%2C+x%5E6%29\"$
\n" ); document.write( "(Since x is the base of a logarithm in this problem, we can ignore the negative 6th roots of 64. Bases of logarithms must be positive!) Since $\"2%5E6+=+64\"$ we get 1/2 on the left side:
\n" ); document.write( " $\"1%2F2+=+x\"$ \n" ); document.write( "

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