Question: Powers of 4

Comment on Powers of 4

I don't get how when you take common 4^(x-2) then you get 4^2 inside the bracket. plz elaborate.
gmat-admin's picture

Great question! To answer it, consider these more straightforward factoring examples:
k^5 - k^3 = k^3(k^2 - 1)
m^19 - m^15 = m^15(m^4 - 1)
In the first example, we factored out k^3 since 3 is the smaller of the two exponents.
In the second example, we factored out m^15 since 15 is the smaller of the two exponents.
With 4^x and 4^(x-2), the exponents are x and (x-2). Here, the smaller exponent is (x-2), so we'll factor out the 4^(x-2) to get: 4^(x-2) times [something] = 4^x - 4^(x-2)
In this case, we must recognize that [4^(x-2)](4^2) = 4^x, since we must add the exponents, and (x-2) + 2 = x

Does that help?

1Narrative's picture

Hi, Brent! Tried a different approach concerning the numerator : (Any idea why it doesn't work?)

4^x - ( 4^x )(4^-2) divided by 5

4^x (1 - 1/4^2) divided by 5

4^x (15/16)(1/5)

= (4^2x)(4^1)
gmat-admin's picture

You're on the right track, but there's a problem with the right side.
It should be 4^(2x+1) - 4^(2x) = (4^2x)[4^1 - 1]

So, we get: (4^x)(15/16)(1/5) = (4^2x)[4^1 - 1]

Simplify: (4^x)(3/16) = (4^2x)[3]

Divide both sides by 3 to get: (4^x)(1/16) = 4^(2x)

Rewrite 1/16 to get: (4^x)[4^(-2)] = 4^(2x)

Simplify left side: 4^(x-2) = 4^(2x)

So, x-2 = 2x

Solve: x = -2

1Narrative's picture


Thanks so much. That gave me a sleepless night - lost a bit of confidence right there.


I don't understand how you factorized as the first step? Can you explain as if you were explaining to a layman? (I did see the first comment but still doesnt make sense) :(
gmat-admin's picture

The key is that we always factor out the term with the SMALLEST exponent.

For example, we take x^8 + x^5 + x^11 and factor out x^5.
We get: x^8 + x^5 + x^11 = x^5(x^3 + 1 + x^6)

With 4^x and 4^(x-2), the exponents are x and (x-2).
Here, the smaller exponent is (x-2), so we'll factor out the 4^(x-2)
At this point, the tricky part is determining what goes in the brackets.
When we multiply powers we ADD exponents.
So, [4^(x-2)][4^2] = 4^(x-2 + 2) = 4^x

So, 4^x - 4^(x-2) = [4^(x-2)](4^2 - 1)

Here are three more examples:
5^(x+1) - 5^(x+8) = 5^(x+1)[1 - 5^7]
2^(x+2) + 2^(x-1) = 2^(x-1)[2^3 + 1]
7^(x-3) + 7^(x-9) = 7^(x-9)[7^6 + 1]

Does that help?


HI Brent, I was also struggling with the factoring in the video so I came to the comments to see that others had already asked. I think I've grasped it better, but in this first example here:

5^(x+1) - 5^(x+8) = 5^(x+1)[1 + 5^(x+7)]

Shouldn't it be 5^(x+1) - 5^(x+8) = 5^(x+1)[1 - 5^(7)]?

I multiplied this to: [5^(x+1)] * [5^(x+7)] = 5^(2x+8) ?
gmat-admin's picture

You're absolutely right, Kameron. Good catch!
I've edited my response above.

Hi Brent,

I solved the below question by factoring.

The value (2^-14 + 2^-15 + 2^-16 + 2^-17)/5 is how many times the value of 2^-17?
2^-14(1 + 2^-1 + 2^-2 + 2^-3)/5
2^-14 (1 + 1/2 + 1/4 + 1/8)/5
2^-14 (15/8)/5
2^-14 * 3/8
2^-14 * 3 * 2^-3
2^-17 * 3
So it is 3 times the value of 2^-17
gmat-admin's picture

Question link:

That approach works perfectly. Nice work!

Interesting. Although I did solved but I took out 4^x out and had to solve the denominator which took me fraction of more time. But as your solution I should have taken the negative one out.. Interesting how little things could save time. Next time I will try to notice this one too.

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