MGMAT Test Q

Hi

this Did take a while !

the answer is -9

here how i did it

the basic intention i had is to elliminate the x exponent value from denominator

so the question can be re written as :

-->[3^12x * 3^6]* [(1/2^ 3x*3^8x)]*[(2^2*3^2)^x]*[3^12/3^4x]

(p.s- exponents in big bracket represenst every terms)

-->[3^12x*3^6*2^2x*3^2x*3^12]/[2^2x*3^8x*3^4x]

canceling and simplifying


we get --> 3^18*3^2x=1

3^2x=1/3^18

3^2x=3^-18

ie 2x=-18

x= -9

Hope this helps

vishu

smallsorrow wrote:27^4x+2 * 162^-2x * 36^x * 9^6-2x=1, then what is the value of x?

I can not understand the equation. I possible pl add brackets.

A) -9
B) -6
C) 3
D) 6
E) 9

Again a nasty exponent problem.

OA follows

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

here with brackets...

(27^4x+2)* (162^-2x) * (36^x) * (9^6-2x)

break them seperately to make it simple...

(27^4x+2)= (3^2)^4x+2
-->3^12x * 3^6---(a)

(162^-2x) where 162 can be written as 2*3^4
and (2*3^4)-4x is 1/(2*3^4)2x
-->1/2^2x*3^8x----(b)

(36^x) is wrritten as 2^2*3^2
-->2^2x*3^2x--(c)

(9^6-2x) .. where it is wrritten as 3^2(6-2x)
-->3^(12-4x)
-->3^12/3^4x---(d)

--> (3^12)^-2x
-->1/3^24x----(d)

now put the values back
-->a*c/b*d

numerator and deno simpolies to further step explained above


HTH

Vishu

Wats the OA????

x should be -9

27^4x+2)* (162^-2x) * (36^x) * (9^6-2x)=1
(9*3) ^ 4x + 2 * (9*9*2) ^ -2x * (9*4) ^x * 9 ^6 * 9 ^ -2x
I got all bases to 9

9 ^(4x+2+2x+1-2x-2x+x+6-2x) * 2 ^ -2x * 2 ^ -2x = 1

9 ^ (x+9) = 9 ^ 0
x+9 = 0
x = -9

Corrected the typo in bold

(27^4x+2)* (162^-2x) * (36^x) * (9^6-2x)=1
(9*3) ^ 4x + 2 * (9*9*2) ^ -2x * (9*4) ^x * 9 ^6 * 9 ^ -2x
I got all bases to 9

9 ^(4x+2+2x+1-2x-2x+x+6-2x) * 2 ^ -2x * 2 ^ 2x = 1

9 ^ (x+9) = 9 ^ 0
x+9 = 0

x = -9