Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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
Substituting 2 for x, we get
2(2) + 4(2) + 2 = k 4 + 8 + 2 = k 14 = k
Answer : E
Last edited by meshell on 23 Aug 2011, 18:55, edited 1 time in total.