solving for x when exponents involved.

[semidevil]

If 4^x + 4^(-x) = 2, which of the following is the value of x?
a -1
b -1/2
c 0 <---- OA
d 1/2
e 1

the easiest way is to plug it in, but how do you solve this algebraically?

my attempt:
-match the base:
4^x + 4^(-x) = 2
==> 2^2^x + 2^2^(-x) = 2^1.
-drop the base
==> 2x + 2(-x) = 1.
==> 2x -2x = 1.
==> 0x = 1

where did I go wrong?

[Jose Ferreira]

Hi,

A few comments on this question:

1. Plugging in the answer choices is certainly the easiest approach

2. Be VERY careful when "dropping the base." This is ONLY legal when you have exactly one term on each side of the equation. For example, if you have:
2^(3 - x) = 2 ^(2x + 9), you can simplify to (3 - x) = (2x + 9) and then solve for x.
When you have an equation with terms that are being added, you CANNOT do this. For an illustration of the fact that this is illegal, consider:
4^1 + 4^1 + 4^1 + 4^1 = 4^2 is TRUE
but
1 + 1 + 1 + 1 = 2 is FALSE

3. One algebraic approach here requires substitution of a relatively unusual type for the GMAT. Having said that, if you are faced with a very tough algebra question and are looking for a way out, it is worth keeping in mind.

The first step is to re-write the equation as 4^x + 1/(4^x) = 2. Note that for this approach, it doesn't matter if you use 4^x or 2^[2x]; the result is the same.

Now, substitute a new variable for 4^x. a = 4^x. This is a legal operation. All we are doing is calling 4^x by another name. Since 4^x is never zero, we won't run into any problems multiplying or dividing by a.

Now, we have a + 1/a = 2. This kind of equation is not uncommon on the GMAT. You typically want to solve it by multiplying through by a to get a quadratic:
a^2 + 1 = 2a, or a^2 - 2a + 1 = 0. Since this is a perfect square, we can factor this to (a - 1)^2 = 0, or a - 1 = 0, or a = 1.
Now, plug 4^x back in for a. We get 4^x = 1. We know this is true when x = 0.

[gmat740]

A more simple solution


Let 4^x = t

so we have t + 1/t = 2

this gives a quadratic equation,

t^2 - 2t + 1 = 0

(t-1)^2 = 0

t=1

but t = 4^x

so,4^x = 1
4^x = 4^0

so compare both sides of the equation

x = 0


We cannot drop the base as you did.

May be for this question you are getting an answer but this might not be the case with other questions


Karan