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Re: A very difficult math [#permalink]
23 Aug 2011, 17:57

Jasonammex wrote:

If x and K are integers and (12^x)(4^2x+1)=(2^k)(3^2), what is the value of K? \((12^x)(4^(2x+1))=(2^k)(3^2)\) ==> \((3^x)(4^x)(4^(2x+1))=(2^k)(3^2)\) ==> \((3^x)(4^(3x+1))=(3^2)(4^(k/2))\)

So \(3^x = 3^2\) ==>\(x = 2\) \(4^(3x+1) = 4^(k/2)\) ==> \(k = 2(3x+1) = 2(3*2+1) = 14\) So OA -E a) 5 b)7 c)10 d)12 e)14

Pls explain me and there gotta be a simple way to solve this?

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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