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## Comment on

Powers of 4## I don't get how when you take

## Great question! To answer it,

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?

## Hi, Brent! Tried a different

LEFT

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

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

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

RIGHT

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

## You're on the right track,

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

## Wow!

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

Thanks.

## I don't understand how you

## The key is that we always

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?

Cheers,

Brent

## HI Brent, I was also

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) ?

## You're absolutely right,

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

## Question link: https:/

Question link: https://gmatclub.com/forum/the-value-of-2-14-2-15-2-16-2-17-5-is-130682....

That approach works perfectly. Nice work!

## Interesting. Although I did