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Re: A very difficult math [#permalink]
23 Aug 2011, 17:59
\((12^x)(4^(2x+1)) = (2^k)(3^2)\)
we want to get the bases to match, so we can then solve for k in the exponent.
12 can be broken down as 3 x 2 x 2. As the rules for exponents go, if the exponents are the same then you can perform the operation as normal with the bases so \(12^x = (2^x)(2^x)(3^x)\). you also want the 4 down to a base of 2, and square root of 4 is 2 so \(2^2=4\), so again using another exponent rule : \((2^2)^(2x+1)\) we multiply the exponents, and get \(2^(4x+2)\)
\((2^2x)(3^x)(2^(4x+2)) = (2^k)(3^2)\)
Now we've got two terms with a base of 2, so we multiply them : which means we add the exponents:
\((2^(2x+4x+2))(3^x) = (2^k)(3^2)\)
Now we can compare both sides, since the bases are the same. So we can assume, since \(3^x = 3^2\), then \(x=2\) and \(2^(2x+4x+2) = 2^k\), therefore \(2x+4x+2 = k\)
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