document.write( "Question 1139283: Solve:\r
\n" ); document.write( "\n" ); document.write( "1) 5^2X^2^+^3X=25\r
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\n" ); document.write( "\n" ); document.write( "2) 3^-X=7^X+^2\r
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\n" ); document.write( "\n" ); document.write( "3) 250000=100000(1+.06/12)^12(t)
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Algebra.Com's Answer #757073 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "1) 5^2X^2^+^3X=25

\n" ); document.write( "The sequence of characters \"X^2+^\" has no meaning.
\n" ); document.write( "The sequence of characters \"+^3\" has no meaning.

\n" ); document.write( "2) 3^-X=7^X+^2

\n" ); document.write( "The sequence of characters \"+^2\" has no meaning.

\n" ); document.write( "3) 250000=100000(1+.06/12)^12(t)

\n" ); document.write( "The equation as you show it is solved by ordinary arithmetic....

\n" ); document.write( "\"250000=100000%281%2B.06%2F12%29%5E12%28t%29\"

\n" ); document.write( "The value of t is 250000, divided by the product of 100000 and (1+.06/12)^12.

\n" ); document.write( "Surely you meant

\n" ); document.write( "3) 250000=100000(1+.06/12)^(12t)

\n" ); document.write( "This is now an equation for finding the number of years t that it takes 100000 to increase to 250000 at 6% compounded monthly.

\n" ); document.write( "\"250000=100000%281%2B.06%2F12%29%5E%2812t%29\"
\n" ); document.write( "\"2.5+=+%281.005%29%5E%2812t%29\"

\n" ); document.write( "The variable is in the exponent; you need to take logs of both sides.

\n" ); document.write( "\"log%28%282.5%29%29+=+%2812t%29%2Alog%28%281.005%29%29\"
\n" ); document.write( "\"12t+=+log%28%282.5%29%29%2Flog%28%281.005%29%29\"

\n" ); document.write( "Use a calculator....
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