Exponent Issue

2^x - 2^(x-2) = 3 * 2^17

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

now
2^(x-2) * 3=3 * 2^17
x-2=17=>19

D

I factored it a bit different than the poster above me, but it's up to you what's easier for you.

problem: 2^x - 2^(x-2) = 3 * 2^17

Working just the left side of the equation:

2^x - 2^(x-2)

factor out 2^(x-2), which results in 2^(x-2) * (2^2 - 1) , which equals 2^(x-2)*3



going back to the full formula ... 2^(x-2) * 3 = 3* 2^17 .. 3's on both sides, so it's just 2^(x-2) = 2^17 ... solving for x is x-2=17 , x = 19 ... D

jpjp wrote:I factored it a bit different than the poster above me, but it's up to you what's easier for you.

problem: 2^x - 2^(x-2) = 3 * 2^17

Working just the left side of the equation:

2^x - 2^(x-2)

factor out 2^(x-2), which results in 2^(x-2) * (2^2 - 1) , which equals 2^(x-2)*3



going back to the full formula ... 2^(x-2) * 3 = 3* 2^17 .. 3's on both sides, so it's just 2^(x-2) = 2^17 ... solving for x is x-2=17 , x = 19 ... D

Excellent job!

Now let's be sneaky:

we see two exponents on the left side: x and (x-2)
we see one exponent on the right side: 17

We know that after factoring, either x or (x-2) will turn out to be 17.

So, either x=17 or x=19; narrow it down to B and D.

Thinking it through, if we factor out 2^(x-2), well get integers (per jpjp); if we factor out 2^x, we'll get fractions (per outreach). We know the final answer has integers, therefore it must be the (x-2) that works out to 17... choose (D).

Can you please explain the following step - I'd like to better understand the shortcut as I usually am tight on time

"factor out 2^(x-2), which results in 2^(x-2) * (2^2 - 1) , which equals 2^(x-2)*3 "

Appreciate your help!

gmatassistance wrote:Can you please explain the following step - I'd like to better understand the shortcut as I usually am tight on time

"factor out 2^(x-2), which results in 2^(x-2) * (2^2 - 1) , which equals 2^(x-2)*3 "

Appreciate your help!

Its a bit hard to explain, but I'll do my best.

The factoring is based on the principle of [a^b] - [a^(b-1)] = [a^b] * [1 - a^(-1)] = [a^(b-1)] * [a - 1]

or if it's easier for you to look at real number examples , the principle is 5^3 - 5^2 = 5^3 * (1 - 5^-1) = 5^2 * (5-1)

soo.. using that principle to factor our problem at hand..

The two numbers you have on that side of the equation are as follows
2^x minus 2^(x-2) , where blue is the first term, green is the second, and red is our factor

They have the same base of 2, so you should recognize that you can factor. If possible, you want to factor where the variable is factored out completely.

so factoring out 2^(x-2), you get 2^2 minus 1

this is because if you factor out 2^(x-2) from 2^x , you have 2^[x-(x-2)] = 2^2 = 4

and

when you factor out 2^(x-2) from 2^(x-2), you get 2^[(x-2)-(x-2)] = 2^0 = 1

therefore
2^(x-2) * (4-1)

hope that helps!

gmatassistance wrote:Can you please explain the following step - I'd like to better understand the shortcut as I usually am tight on time

"factor out 2^(x-2), which results in 2^(x-2) * (2^2 - 1) , which equals 2^(x-2)*3 "

Appreciate your help!

When we factor out, we're using the multiplication rule of exponents:

x^a * x^b = x^(a+b)

We need to factor out because you can't add or subtract terms with variables unless they have the same exponent.

For example, if we have:

2(x^2) + 3(x^3),

there's no simple way to combine those terms. However, if we have:

2(x^2) + 3(x^2)

we simply add the coefficients and get:

5(x^2).

So, if we want to add or subtract terms with variables, we need to equalize the exponents.

Before getting to this question, let's look at a simpler example:

2^8 - 2^6 = ?

We think to ourselves: we need to equalize the exponents. As a general rule, we always simplify to the smaller of the powers.

So, how can we express 2^8 with an exponent 6?

Well, using the multiplication rule noted up top, we can say that:

2^8 = 2^2 * 2^6 = 4 * 2^6

Plugging back into the question:

2^8 - 2^6 = 4(2^6) - 1(2^6) = 3(2^6)

(I rewrote 2^6 as 1(2^6) just to clarity the coefficient of that term.)

Now back to the question in this thread:

2^x - 2^(x-2)

(x-2) is the smaller exponent, so let's make that our main exponent for the expression.

So, we need to rewrite 2^x with a power of (x-2) instead of just x.

Well, using the multiplication rule of exponents:

2^x = 2^(x-2+2) = 2^(x-2) * 2^2 = 4(2^(x-2))

Substituting into the original expression:

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

Thanks so much for taking the time out for the detailed explanations above. The more I stare at it the more sense it makes to me - great trick!