If [4^x - 4^(x-2)]/5 = 4^(2x+1) - 4^(2x), then x = ?

-3.5
-3
-2.5
-2
-1.5

We can PLUG IN THE ANSWERS, which represent the value of x.
Exponents on the GMAT tend to be INTEGER VALUES.
Thus, the correct answer choice is likely B or D.

Strategy:
Factor out the term with the SMALLER EXPONENT.

D: -2
If we plug x=-2 into [4^x - 4^(x-2)]/5 = 4^(2x+1) - 4^(2x), we get:
(4¯² - 4¯�)/5 = 4¯³ - 4¯�.

On each side, factor out 4¯�:
4¯�(4² - 1)/5 = 4¯�(4 - 1)

Cancel out the terms in red and simplify the terms in blue:
15/5 = 3

3 = 3.

Success!

The correct answer is D.

Algebraic solution:

[4^x - 4^(x-2)]/5 = 4^(2x+1) - 4^(2x)

On the left side, factor out 4^(x-2).
On the right side, factor out 4^(2x):
4^(x-2)(4² - 1)/5 = 4^(2x)(4 - 1)

Simplify the terms in blue:
4^(x-2)(3) = 4^(2x)(3)

Cancel out the terms in blue:
4^(x-2) = 4^(2x)

Since the bases match, so must the exponents.
Thus:
x-2 = 2x
-2 = x.

The correct answer is D.

Ok. I think i get it.

In the algebraic approach, when you factored 4^(x-2) from the equation of the left and the answer was 4^2. The approach was to:
1) Take the smaller exponent which is: 4^(x-2)
2) When you factor 4^(x-2) from 4^x then you need to find a value that converts 4^(x-2) to 4^x which is 4^2. Because 4^x-2(+2) makes 4^x. By adding 4^+2 we converted 4^x-2 to 4x. Therefore, 4^2 remains after factoring 4^x-2 from 4^x. Is this correct?[/b]

shahfahad wrote:Ok. I think i get it.

In the algebraic approach, when you factored 4^(x-2) from the equation of the left and the answer was 4^2. The approach was to:
1) Take the smaller exponent which is: 4^(x-2)
2) When you factor 4^(x-2) from 4^x then you need to find a value that converts 4^(x-2) to 4^x which is 4^2. Because 4^x-2(+2) makes 4^x. By adding 4^+2 we converted 4^x-2 to 4x. Therefore, 4^2 remains after factoring 4^x-2 from 4^x. Is this correct?[/b]

Correct!

But in this question:

https://www.beatthegmat.com/advanced-qua ... 79853.html

Brent factored (5^y) from 5^x - 5^y and the result was (5^y)[5^(x-y) - 1]

I am having trouble understanding this part. How can you factor 5^y from 5^x. Then dont have anything in common. And even if you do, the result should have been (according to me which i know is incorrect obviously because Brent cant be incorrect) (5^y)(5^x*y-1) rather than (5^y)[5^(x-y) - 1]. Why -y? Please help.

Also, can i get a link to any article/website/video lecture or anything that explains in detail the factoring of numbers and especially variables? Any good resource? Because i am struggling with the factoring of variables.

Hi Brent. Can you please explain the factoring of variables. I did not understand how you factored the above equation.

When you factored (5^y) from 5^x - 5^y and the result was (5^y)[5^(x-y) - 1]

I am having trouble understanding this part. How can you factor 5^y from 5^x. Then dont have anything in common. And even if you do, the result should have been (5^y)(5^x*y-1) rather than (5^y)[5^(x-y) - 1]. Why -y? Please help.

shahfahad wrote:When you factored (5^y) from 5^x - 5^y and the result was (5^y)[5^(x-y) - 1]

I am having trouble understanding this part. How can you factor 5^y from 5^x. Then dont have anything in common. And even if you do, the result should have been (5^y)(5^x*y-1) rather than (5^y)[5^(x-y) - 1]. Why -y? Please help.

Maybe think of it this way: when you're factoring something out of an expression, you're dividing each term by the thing you've factored out. So 2x - 4 becomes 2(x-2), because both the '2x' term and the '4' term have been divided by 2. Similarly, 4^5 - 4^3, becomes 4^3(4^2 - 1) because you've divided each term by the 4^3 that you've factored out.

But using this logic, we can factor out anything we want. If you wanted to, you could factor a '3' out of 2x-4 and call the expression 3 * [(2x/3) - 4/3).] Even though '3' isn't necessarily a factor of either term, we're effectively multiplying by 3/3, and so we haven't changed the expression. (And you can see that if we distribute the '3', then 3 * [(2x/3) - 4/3), will become 2x - 4. The expressions are the same.

Similarly, if we started with 5^x - 1 we can factor out whatever we want. You could factor out a 2, and call the expression 2[(5^x)/2 - (1/2). There's no real reason to do this, but it's mathematically legal. And we could start with 5^x - 1 and then factor out a 5^y, to get 5^y*[(5^x)/(5^y) - 1/(5^y).] We just divided each term by 5^y. (Note that (5^x)/(5^y) = 5^(x-y). I'm using (5^x)/(5^y) for simplicity's sake.) Note, also, that if we distribute the 5^y, then 5^y*[(5^x)/5^y) - 1/(5^y), will become 5^x - 1. So when we factor something out, we're really just dividing each term by the thing we've factored.

Cool. That was helpful. Thank you all for your help.