document.write( "Question 744487: 1. Solve the following equations
\n" ); document.write( "a) 5^(2x) + 5^(x) - 6 = 0 (hint let y = 5^(x)\r
\n" ); document.write( "\n" ); document.write( "Can you please help me out? Thanks so much in advance:)
\n" ); document.write( "Can you please show all the steps it would really help me understand:)
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Algebra.Com's Answer #453344 by nerdybill(7384)\"\" \"About 
You can put this solution on YOUR website!
a) 5^(2x) + 5^(x) - 6 = 0 (hint let y = 5^(x)
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\n" ); document.write( "We Let y = 5^x
\n" ); document.write( "if so, we can rewrite:
\n" ); document.write( "5^(2x) + 5^(x) - 6 = 0
\n" ); document.write( "AS
\n" ); document.write( "y^2 + y - 6 = 0
\n" ); document.write( "now, we factor :
\n" ); document.write( "(y+3)(y-2) = 0
\n" ); document.write( "y = {-3, 2)
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\n" ); document.write( "But, we still need to find x:
\n" ); document.write( "since
\n" ); document.write( "y = 5^x
\n" ); document.write( "we substitute each value of y to find x:
\n" ); document.write( "-3 = 5^x
\n" ); document.write( "\"log%285%2C-3%29+=+x+\"
\n" ); document.write( "since we can't take the log of a negative value we throw out this solution.
\n" ); document.write( ".
\n" ); document.write( "2 = 5^x
\n" ); document.write( "\"log%285%2C2%29+=+x+\"
\n" ); document.write( "\"log%282%29%2Flog%285%29+=+x+\"
\n" ); document.write( "0.431 = x
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