If \(4^{(2x+1)} + 4^{(x+1)} = 80\), what is the value of x ?

\(4*4^{2x} + 4*4^{x} —80 = 0\)
\(4^{2x} + 4^{x} —20 =0\)
\((4^{x} —4)(4^{x} +5) = 0\)
—> \(4^{x} —4 =0\)
\(4^{x} = 4^{1}\)—> \(x= 1\)

The answer is C

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Hi lacktutor,

Can you explain how you factorize this?
Thanks!

Hi InaKi20,

Let’s say that \(4^{x} = a\)
—>\( a^{2} + a—20 = 0\)
(Simple quadratic equation)
(a—4)(a+ 5) = 0
a= 4 and a= —5
—>\( 4^{x} =4\) and \(4^{x} =—5\)
(\(4^{x}\) —always greater than zero)

—> \(4^{x} =4\) —> x= 1

Hope it helps

If 4^(2x+1) + 4^(x+1) = 80, what is the value of x?
factor out the smallest (4^(x+1))
4^(x+1)•[4^x +4^0] =80
4^(x+1)•[4^x+1] = 80
4^(x+1)•[4^x+1]= 4^2•5
For the above to be equal
4^(1+1)•(4^1+1) = 4^2•5
4^2•5 = 4^2•5

x=1 (Answer C)

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Solution:

Simplifying, we have:

(4^2x)(4) + (4^x)(4) = 80

4(4^2x + 4^x) = 80

4^2x + 4^x = 20

Just by looking at the answer choices, we see that x must be 1 since 4^2 + 4^1 = 20.

Alternate Solution:

Simplifying, we have:

4^(x+1) * (4^x + 1) = 4^2 * 5

We see that if x = 1, we have 4^(x+1) = 4^2 and 4^x + 1 = 4 + 1 = 5. So x must be 1.

Answer: C
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Just plug numbers quickly to find c. That should not take more than 20 sec

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STRATEGY: Upon reading any GMAT Problem Solving 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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