document.write( "Question 278548: Find each value of x.\r
\n" ); document.write( "\n" ); document.write( " $\"2%5Elog2%5E5=x\"$ . \n" ); document.write( "

Algebra.Com's Answer #202948 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"2%5Elog2%5E5+=+x\"$
\n" ); document.write( "If this is what the problem really looks like then, since $\"2%5E5+=+32\"$ , this simplifies to:
\n" ); document.write( " $\"2%5Elog%28%2832%29%29=x\"$
\n" ); document.write( "Now we use our calculators to find log(32) and then use the calculators again to raise to to that power.

\n" ); document.write( "However, I suspect the the problem is actually:
\n" ); document.write( " $\"2%5Elog%282%2C+%285%29%29+=+x\"$
\n" ); document.write( "This equation is extremely easy if you understand what logairthms are. Logarithms are exponents. In general, $\"log%28a%2C+%28b%29%29\"$ represents the exponent you put on a which results in b. This specific logarithm, $\"log%282%2C+%285%29%29\"$ , represents the exponent you put on 2 to get 5. And where do we see $\"log%282%2C+%285%29%29\"$ ? We see it as an exponent on 2! So $\"2%5Elog%282%2C+%285%29%29\"$ , by the definition of logarithms, is 5!
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