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

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Math Expert

Joined: 02 Sep 2009

Posts: 61548

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

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28 Jan 2020, 02:19

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Question Stats:

85% (01:45) correct

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

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

Spoiler: OA

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Joined: 25 Jul 2018

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

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28 Jan 2020, 02:30

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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Joined: 19 Jan 2020

Posts: 19

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

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28 Jan 2020, 08:49

lacktutor wrote:

Hi lacktutor,

Can you explain how you factorize this?
Thanks!

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Joined: 25 Jul 2018

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

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28 Jan 2020, 11:31

InaKi20 wrote:

lacktutor wrote:

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

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

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28 Jan 2020, 15:09

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

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02 Feb 2020, 09:48

Bunuel 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

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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