document.write( "Question 670878: what is the solution to the equation log4 of 4x + 2log4 of x = 4a
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Algebra.Com's Answer #417172 by Theo(13342) $\"\"$ $\"About$

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the answer is x = 4
\n" ); document.write( "here's how.
\n" ); document.write( "your equation is:
\n" ); document.write( "log4(4x) + 2log4(x) = 4
\n" ); document.write( "since a*log(b) = log(b^a), this equation becomes:
\n" ); document.write( "log4(4x) + log4(x^2) = 4
\n" ); document.write( "since log(a) + log(b) = log(a*b), this equation becomes:
\n" ); document.write( "log4(4x * x^2) = 4
\n" ); document.write( "simplify this to get:
\n" ); document.write( "log4(4x^3) = 4
\n" ); document.write( "since logb(x) = y if and only if b^y = x, this equation becomes:
\n" ); document.write( "4^4 = 4x^3
\n" ); document.write( "since 4^4 = 256, this equation becomes:
\n" ); document.write( "256 = 4x^3
\n" ); document.write( "divide both sides of this equation by 4 to get:
\n" ); document.write( "64 = x^3
\n" ); document.write( "take the third root of both sides of this equation to get:
\n" ); document.write( "x = 4
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