document.write( "Question 285384: $\"25%5Ex=1%2F5\"$ \n" ); document.write( "

Algebra.Com's Answer #206991 by jsmallt9(3758) $\"\"$ $\"About$

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If you understand exponents well you can just look at this and know what x must be. 1/5 is the reciprocal of 5 and 5 is the square root of 25. Since the exponent for reciprocals is -1 and the exponent for square roots is 1/2, $\"x+=+-1%2F2\"$ !

\n" ); document.write( "If you don't get exponents that well, you can use logarithms. Any base of logarithm will do but the best ones would be 5 (because both 25 and 1/5 are powers of 5), 10 or e (because your calculator \"knows\" these bases.

\n" ); document.write( "Using base 5 logarithms:
\n" ); document.write( " $\"log%285%2C+%2825%5Ex%29%29+=+log%285%2C+%281%2F5%29%29\"$
\n" ); document.write( "Using a property of logarithms, $\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"$ , we can move the exponent of the logarithm on the left side out in front:
\n" ); document.write( " $\"x%2Alog%285%2C+%2825%29%29+=+log%285%2C+%281%2F5%29%29\"$
\n" ); document.write( "Since $\"25+=+5%5E2\"$ and $\"1%2F5+=+5%5E%28-1%29\"$ , the logarithms are 2 and -1, respectively:
\n" ); document.write( " $\"x%2A2+=+-1\"$
\n" ); document.write( "Then we divide by 2:
\n" ); document.write( " $\"x+=+-1%2F2\"$

\n" ); document.write( "Using base 10 logarithms instead:
\n" ); document.write( " $\"log%28%2825%5Ex%29%29+=+log%28%281%2F5%29%29\"$
\n" ); document.write( " $\"x%2Alog%28%2825%29%29+=+log%28%281%2F5%29%29\"$
\n" ); document.write( " $\"x+=+log%28%281%2F5%29%29%2Flog%28%2825%29%29\"$
\n" ); document.write( "Then we get out our calculator and find the two logarithms and then divide them. You should get -0.5 (which is -1/2 in decimal form) or some number very, very close to -0.5. (Calculators use decimal approximations for logarithms so there may be a small rounding error. This is why it is preferable to avoid calculators when you can.) \n" ); document.write( "

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