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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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25% (medium)

Question Stats:

85% (01:45) correct 15% (01:33) wrong based on 47 sessions

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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

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Director
Joined: 25 Jul 2018
Posts: 578
If 4^(2x + 1) + 4^(x + 1) = 80, what is the value of x ?  [#permalink]

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28 Jan 2020, 02:30
2
If $$4^{(2x+1)} + 4^{(x+1)} = 80$$, what is the value of x ?

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

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

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

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Hi lacktutor,

Can you explain how you factorize this?
Thanks!
Director
Joined: 25 Jul 2018
Posts: 578
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
1
InaKi20 wrote:
lacktutor wrote:
If $$4^{(2x+1)} + 4^{(x+1)} = 80$$, what is the value of x ?

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

Posted from my mobile device

Hi lacktutor,

Can you explain how you factorize this?
Thanks!

Hi InaKi20,

Let’s say that $$4^{x} = a$$
—>$$a^{2} + a—20 = 0$$
(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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Posts: 468
Location: Ghana
Concentration: Finance, Real Estate
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

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Target Test Prep Representative
Status: Founder & CEO
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Joined: 14 Oct 2015
Posts: 9535
Location: United States (CA)
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.

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Intern
Joined: 06 Jan 2018
Posts: 8
Re: If 4^(2x + 1) + 4^(x + 1) = 80, what is the value of x ?  [#permalink]

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02 Feb 2020, 10:04
Just plug numbers quickly to find c. That should not take more than 20 sec

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Re: If 4^(2x + 1) + 4^(x + 1) = 80, what is the value of x ?   [#permalink] 02 Feb 2020, 10:04
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