document.write( "Question 288641: 25^x-1=125^4x \n" ); document.write( "

Algebra.Com's Answer #209311 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"25%5E%28x-1%29=125%5E%284x%29\"$
\n" ); document.write( "The simplest solution to this is based on recognizing that both 25 and 125 are powers of 5. So we can express both side as powers of 5:
\n" ); document.write( " $\"%285%5E2%29%5E%28x-1%29=%285%5E3%29%5E%284x%29\"$
\n" ); document.write( "Using the proper rule for exponents (i.e. multiply them) we get:
\n" ); document.write( " $\"5%5E%282x-2%29=5%5E%2812x%29\"$
\n" ); document.write( "With both sides now being powers of 5, they can only be equal if the exponents are the same:
\n" ); document.write( "2x-2 = 12x
\n" ); document.write( "Solving this:
\n" ); document.write( "-2 = 10x
\n" ); document.write( "-2/10 = x
\n" ); document.write( "-1/5 = x

\n" ); document.write( "If you don't recognize that 25 and 125 are powers of 5 we can still use logarithms:
\n" ); document.write( " $\"25%5E%28x-1%29=125%5E%284x%29\"$
\n" ); document.write( " $\"log%28%2825%5E%28x-1%29%29%29=log%28%28125%5E%284x%29%29%29\"$
\n" ); document.write( " $\"%28x-1%29log%28%2825%29%29=4x%2Alog%28%28125%29%29\"$
\n" ); document.write( " $\"x%2Alog%28%2825%29%29-log%28%2825%29%29=4x%2Alog%28%28125%29%29\"$
\n" ); document.write( " $\"-log%28%2825%29%29+=+4x%2Alog%28%28125%29%29+-+x%2Alog%28%2825%29%29\"$
\n" ); document.write( " $\"-log%28%2825%29%29+=+x%2A%284log%28%28125%29%29+-+log%28%2825%29%29%29\"$
\n" ); document.write( " $\"%28-log%28%2825%29%29%29%2F%284log%28%28125%29%29+-+log%28%2825%29%29%29+=+x\"$
\n" ); document.write( "We can use our calculators on this. We should get -0.2 (or some decimal very close to it). 0.2 is the decimal for 1/5 so the answer is still the same as before. \n" ); document.write( "

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