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Re: If 4^(2x + 1) + 4^(x + 1) = 80, what is the value of x ? [#permalink]
1
Carcass wrote:
If \(4^{(2x + 1)} + 4^{(x+1)} = 80\), what is the value of x ?

(A) -5
(B) 0
(C) 1
(D) 4
(E) 5



STRATEGY: Upon reading any GRE Multiple Choice question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can easily test the answer choices.
In fact, if I apply a little bit of number sense, I can quickly rule out answer choices D and E, since \(4^{(2x + 1)}\) and \(4^{(x+1)}\) would each evaluate to be much greater than 80.
From here, I'd typically give myself up to 20 seconds to identify a faster approach, but I can already see that testing the answer choices will be super fast.

Let's start by testing answer choice B, \(x = 0\).

When we plug \(x = 0\) into the given equation we get: \(4^{(2(0) + 1)} + 4^{(0+1)} = 80\)

Simplify to get: \(4^{1} + 4^{1} = 80\)

Evaluate to get: \(4 + 4 = 80\). Doesn't work!

So, we can eliminate answer choice B .
More importantly, we can see that we need the exponents to be bigger in order to get a sum of 80.

So let's now test answer choice C, \(x = 1\).

We get: \(4^{(2(1) + 1)} + 4^{(1+1)} = 80\)

Simplify to get: \(4^{3} + 4^{2} = 80\)

Evaluate to get: \(64 + 16 = 80\)

Works!!

Answer: C
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Re: If 4^(2x + 1) + 4^(x + 1) = 80, what is the value of x ? [#permalink]
\(4^{(2x + 1)} + 4^{(x+1)}\) = 80
=> \(4^{2x} * 4^1 + 4^x * 4^1\) = 80
=> \((4^x)^2 * 4 + 4^x * 4\) = 4*20

Dividing both the sides by 4 we get
=> \((4^x)^2 + 4^x\) = 20

Let \(4^x\) = a we get
=> \(a^2 + a - 20 = 0\)
=> \(a^2 + 5a - 4a - 20 = 0\)
=> a*(a + 5) - 4*(a + 5) = 0
=> (a + 5) * (a - 4) = 0
=> a = 4, -5

=> \(4^x\) = 4, -5
But \(4^x\) cannot be negative

=> \(4^x\) = 4 = \(4^1\)
=> x = 1

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Exponents

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Re: If 4^(2x + 1) + 4^(x + 1) = 80, what is the value of x ? [#permalink]
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