document.write( "Question 1098778: 5^x=3-5^(2x)\r
\n" ); document.write( "\n" ); document.write( "Please help! Sorry I've been asking a lot of questions but I have a test tomorrow and we had a supply teacher who didn't know the material. The teacher left handouts without a lesson so I've been figuring things out on my own. More likely I'm really tired and can't concentrate. Sorry for the inconvenience.
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Algebra.Com's Answer #713162 by greenestamps(13215)\"\" \"About 
You can put this solution on YOUR website!

With the \"5^2x\" and \"5^x\" terms in the equation, this can be viewed as a \"quadratic\" equation in which the \"variable\" is 5^x, because \"%285%5Ex%29%5E2+=+5%5E%282x%29\".

\n" ); document.write( "So rewrite the equation as
\n" ); document.write( "\"5%5E%282x%29+%2B+5%5Ex+-+3+=+0\"

\n" ); document.write( "If it helps ease the confusion, you can define a new variable \"u+=+5%5Ex\" and write the equation as \"u%5E2+%2B+u+-+3+=+0\"

\n" ); document.write( "The quadratic does not factor, so use the quadratic formula:
\n" ); document.write( "\"5%5Ex+=+%28-1%2B-sqrt%2813%29%29%2F2\"

\n" ); document.write( "Since 5^x is never negative, we choose the positive solution: \"5%5Ex+=+%28-1%2Bsqrt%2813%29%29%2F2\"

\n" ); document.write( "Then to solve for x, since it is an exponent, we take logs of both sides:
\n" ); document.write( "\"x%2Alog%285%29+=+log%28%28-1%2Bsqrt%2813%29%29%2F2%29\"
\n" ); document.write( "\"x+=+log%28%28-1%2Bsqrt%2813%29%29%2F2%29%2Flog%285%29\"
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