need help on two quant questions

Question 1

Simplify the LHS:

2^x - 2^x-2 = 2^x - 2^x*2^-2
= 2^x*(1-1/4)
= 2^x*(3/4)
= (2^x*2^-2)*3
= 3(2^x-2)

Equating with RHS ==> x-2 = 13
=> x=15

--------------------

Question 2

For each option replace 'x' with '1-x', and you'll find that for option D:

f(1-x) = (1-x)^2(1-1+x)^2
= (1-x)^2(x)^2
= x^2(1-x)^2
===> f(x) = f(1-x)

Hi Prashant,

I still don't understand how you simplified the lefthand side of the first equation of the first problem. How did you get 2^x - 2^x*2-2? If you could clarify this, that'd be awesome. Thanks

2^z = 2^x * 2^y where x + y = z
ex:
2^6 = 2^2 * 2^4 (2+4 = 6)

so 2^(x-2) = 2^x * 2^-2 (x-2 = x-2)

Question 1:
(Use backsolving, plug in the results into x)
for b) (in backsolving i always do this sequence b first if false d or a, because usually the values are ordered and if the result is lower than expected try d, if d is lower e is the answer *usualy* if higher c is the answer)
2^11-2^9=3(2^13)
(divide by 2^13 on both sides and you get)
2^11/2^13-2^9/2^13=3
(using powers properties you get)
2^-2-2^-4=3
(or)
1/(2^2)-1/(2^4)=3 always false (and 3/16 < 3, so try d)

d)
2^15-2^13=3(2^13)
(same approach as above)
2^15/2^13-2^13/2^13=3
2^2-1=3
4-1=3
3=3 always true

D is the answer

Question 2:
as sugested above, plug in 1-x for x in the functions

a)
f(x)=f(1-x)
1-x=1-1+x
2x=1
x=1/2 (only true if x=1/2)

b) x^2=1/2 (following the same procedure)

c) x=1/2

d) x^2*(1-x)^2=(1-x)^2*(1-1+x)^2
x^2*(1-x)^2=(1-x)^2*x^2
(you can already tell that this is always true, but if you can't see it continue)
(divide both sides by x^2*(1-x)^2)
1=1 always true
so D is the answer

no need to bother with e), but you will find that you'll get x=1/2 as well