document.write( "Question 930676: I need your help with this problem please:\r
\n" ); document.write( "\n" ); document.write( "Solve for X: 3^2x+3^x=6\r
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Algebra.Com's Answer #565218 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!
Solve for X: 3^2x+3^x=6
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\"3%5E%282x%29+%2B+3%5Ex+=+6\"
\n" ); document.write( "The easiest and least complicated method is to assign a VARIABLE to \"3%5Ex\". I will assign a, or replace \"3%5Ex\" with a
\n" ); document.write( "Thus, \"3%5E%282x%29+%2B+3%5Ex+=+6\" becomes:
\n" ); document.write( "\"a%5E2+%2B+a+=+6\" ---------- \"3%5E%282x%29\" is the same as \"%283%5Ex%29%5E2\", leading to \"a%5E2\"
\n" ); document.write( "\"a%5E2+%2B+a+-+6+=+0\" ------- Subtracting 6 from each side
\n" ); document.write( "(a - 2)(a + 3) = 0 ------- Factoring quadratic equation to solve for a
\n" ); document.write( "a - 2 = 0 OR a + 3 = 0
\n" ); document.write( "a = 2 OR a = - 3\r
\n" ); document.write( "\n" ); document.write( "a = 2
\n" ); document.write( "Since \"a\" was assigned to \"3%5Ex\", or was used to replace \"3%5Ex\", we can then say that \"a+=+2+=+3%5Ex\"
\n" ); document.write( "\"3%5Ex+=+2\" ------------ Exponential form
\n" ); document.write( "\"log+%283%2C+2%29+=+x\" -------- Logarithmic form
\n" ); document.write( "\"x+=+log+2%2Flog+3\", or \"highlight_green%28x+=+.63093%29\"\r
\n" ); document.write( "\n" ); document.write( "a = - 3
\n" ); document.write( "Since \"a\" was assigned to \"3%5Ex\", or was used to replace \"3%5Ex\", we can then say that \"a+=+-+3+=+3%5Ex\"
\n" ); document.write( "\"3%5Ex+=+-+3\" ---------- Exponential form
\n" ); document.write( "\"log+%283%2C+-+3%29+=+x\" ------ Logarithmic form
\n" ); document.write( "IGNORE the above LOGARITHMIC FORM since a log CANNOT YIELD a value < 0.
\n" ); document.write( "You can do the check!!
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