Problem Solving

If 27^(4x + 2) × 162^(-2x) × 36^x × 9^(6–2x) = 1, then what is the value of x?


-9

-6

3

6

9

Holy crap this is retarded. How in the world are you to solve this in around 2 minutes without f-ing up. I love the GMAT!!!!

HELP
Kaunteya

Hi Kaunteya,

Start off by simplifying all the terms in to their primary factors and then equate the powers of exponents on both sides of equation.

27^(4x + 2) * 162^(-2x) * 36^x * 9^(6–2x) = 1
or (3)^4(4x + 2) * (2*81)^(-2x) * (6*6)^x * 3^2(6–2x) = 1
or (3)^4(4x + 2) * (2)^(-2x) * (3)^4(-2x) * (3*2)^2x × 3^2(6–2x) = 1
or (3)^4(4x + 2) * (2)^(-2x) * (3)^(-8x) * (3)^2x*(2)^2x × 3^2(6–2x) = 1

or (3)^[4(4x + 2) +(-2x)+ 2x+2(6–2x)]* (2)^[(-2x)+2x] = 1
or (3)^[16x +8 -8x + 2x+12–4x]* (2)^[0] = 1
or (3)^[ 6x + 20] = 1 ^ 0 (anything raised to 0 is one)

now compare the powers on both sides
6x+ 20 = 0
or x = -20/6
or x ~ -3


Cross check your question and answer choices? Either one of them is incorrect.

Thanks
Komal

First, you do not need to solve it in 2 minutes. If you prepare well, some of the earlier questions can be solved easily, allowing you to remain CALM when you see a monstrosity like this question!

When you see complicated exponents, it's very likely that there will be some way to simplify.

In this case, the original:

27^(4x + 2) × 162^(-2x) × 36^x × 9^(6–2x) = 1

becomes:

3^3 to the (4x + 2) * 2*3^4 to the (-2x) * 4*3^2 to the x * 3^2 to the (6-2x) = 3^0

the 2^-2x in the second term and the 2^2x in the 3rd term (2^2 to the x) cancel each other, leaving an equation with all terms being 3 raised to a power.

the result is 12x + 6 + (-8x) + 2x + 12 - 4x = 0, which yields

2x + 18 = 0, so x = -9.


If the math looks tough, there's going to be a trick! Be confident, and keep plugging away.

Good luck!
Tony

Hey Tony, thanks for the help on that question. I forgot to raise the 2 to the (-2x). So I was having quite the problem. Your approach was a hell of a lot faster than mine. Good help.

Kaunteya