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If 2^4x = 3,600, what is the value of (2^1-x)^2 ?

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Joined: 24 May 2014
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If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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25 Jun 2014, 19:24
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If $$2^{4x} = 3,600$$, what is the value of $$(2^{(1-x)})^2$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1
Director
Joined: 25 Apr 2012
Posts: 702
Location: India
GPA: 3.21
Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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25 Jun 2014, 21:00
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Reni wrote:
If 2^4x = 3,600, what is the value of (2^1-x)^2 ?

(A)-1/15
(B) 1/15
(C)3/10
(D)-3/10
(E)1

Given 2^4x=1600
We need to find value of 2 ^(1-x)^2-------> Simplify this term

$$(\frac{2}{2^x})^2$$

So we need to find value of $$\frac{4}{2^{2x}}$$

2^4x=3600. Taking a square root we get

2^2x=60

So we get 4/60 or 1/15
Ans is B

Similar question or practice:

given-2-4x-1600-what-is-the-value-of-154486.html#p1236720
if-4-4x-1600-what-is-the-value-of-4-x-161823.html
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Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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10 Jul 2014, 03:35
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$$2^{4x} = 3600$$

$$(2^{2x})^2 = 60^2$$

$$2^{2x} = 60$$ .............. (1)

$$(2^{1-x})^2 = \frac{2^2}{2^{2x}}$$

$$= \frac{4}{60}$$.............. (From 1)

$$= \frac{1}{15}$$

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Intern
Joined: 26 Apr 2016
Posts: 8
Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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08 Jul 2016, 11:57
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$2^{(1-x)^2}$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

Is it $$2^{(1-x)^2}$$ as written in the original post? Or (2^(1-x))^2 as in the answer solutions? The question stem multiplies to be 2^(x^2 - 2x + 1).
Math Expert
Joined: 02 Sep 2009
Posts: 45251
Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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08 Jul 2016, 12:43
DavidFox wrote:
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$2^{(1-x)^2}$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

Is it $$2^{(1-x)^2}$$ as written in the original post? Or (2^(1-x))^2 as in the answer solutions? The question stem multiplies to be 2^(x^2 - 2x + 1).

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It's $$(2^{(1-x)})^2$$
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Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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04 Nov 2016, 20:46
Can someone please explain why we can't do the following:

Break 3600 to 2^4 x (3^2) x (5^2) --> so x=1

Why is this not the correct way to approach?

Math Expert
Joined: 02 Sep 2009
Posts: 45251
Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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05 Nov 2016, 02:08
lawiniecke wrote:
Can someone please explain why we can't do the following:

Break 3600 to 2^4 x (3^2) x (5^2) --> so x=1

Why is this not the correct way to approach?

$$2^{4x} = 2^4*3^2*5^2$$. If x=1, then you'd get that $$1 = 3^2*5^2$$, which is wrong --> $$x \neq 1$$. In fact from $$2^{4x} = 3600$$ it follows that x is some irrational number (approximately 2.9534...), not an integer.
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Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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22 Nov 2017, 12:27
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$(2^{(1-x)})^2$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

Let’s first simplify the expression we want to evaluate. We see that [2^(1-x)]^2 can be simplified as (2 * 2^-x)^2 = 2^2 * 2^-2x = (2^2)/(2^2x)

Thus, if we can determine 2^2x, then we have an answer.

Taking the square root of both sides of the given equation, which is 2^4x = 3600 we have 2^2x = 60; thus:

(2^2)/(2^2x) = 4/60 = 1/15

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Director
Joined: 17 Dec 2012
Posts: 635
Location: India
Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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23 Jan 2018, 18:15
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$(2^{(1-x)})^2$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

2^(4x)= 3600 => x is just less than 3
=>2^((1-x))^2 = approx (2^(1-3))^2= approx 1/(2^4)
Hence B.
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Joined: 09 Mar 2016
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If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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06 May 2018, 03:41
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$(2^{(1-x)})^2$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

pushpitkc hello there, can you please tell what am i doing wrong in my solution and why i am doing it wrong

so we have $$2^{4x} = 3,600$$ (the first thing that came to my mind is that i need to bring both sides to the same base) so here is what i did

$$2^{4x} = 36*10^2$$ (now break down 36 into primes which is $$3^2*2^2$$ see below in next step)

$$2^{4x} = 3^2*2^2*10^2$$ ( here since i want to have the same bases of 2 i do following) turn $$3^2$$ into $$2^3$$ and $$10^2$$ into $$2^{10}$$(see below)

$$2^{4x} = 2^2*2^2*2^{10}$$

$$2^{4x} = 2^{14}$$[ now equate bases

$$4x = 14$$

x= $$\frac{14}{4}$$ --> $$\frac{7}{2}$$ now what

question in conclusion: shoudldnt we always equate bases in similrar question as i did in the above solution:)

thank you
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Joined: 22 Jan 2014
Posts: 155
WE: Project Management (Computer Hardware)
Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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06 May 2018, 04:36
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$(2^{(1-x)})^2$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

2^4x = 3600
taking log both sides
4x log 2 = log 2^4 + log 3^2 + log 5^2
or log 3^2 + log 5^2 = 4 log 2 (x-1)
or log 3 + log 5 = 2 log 2 (x-1)
=> (2^(1-x))^2 = 1/15
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Joined: 26 Feb 2016
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Location: India
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If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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06 May 2018, 05:13
1
dave13 wrote:
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$(2^{(1-x)})^2$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

pushpitkc hello there, can you please tell what am i doing wrong in my solution and why i am doing it wrong

so we have $$2^{4x} = 3,600$$ (the first thing that came to my mind is that i need to bring both sides to the same base) so here is what i did

$$2^{4x} = 36*10^2$$ (now break down 36 into primes which is $$3^2*2^2$$ see below in next step)

$$2^{4x} = 3^2*2^2*10^2$$ ( here since i want to have the same bases of 2 i do following) turn $$3^2$$ into $$2^3$$ and $$10^2$$ into $$2^{10}$$(see below)

$$2^{4x} = 2^2*2^2*2^{10}$$

$$2^{4x} = 2^{14}$$[ now equate bases

$$4x = 14$$

x= $$\frac{14}{4}$$ --> $$\frac{7}{2}$$ now what

question in conclusion: shoudldnt we always equate bases in similrar question as i did in the above solution:)

thank you

Hi dave13

You equate bases whenever it is possible. In cases like these, where
bases can't be equated you will have another options to solve the problem

The reason what you have done is wrong is because
$$2^10 = 1024$$, whereas $$10^2 = 100$$
$$2^3 = 8$$, whereas $$3^2 = 9$$
How can the expressions where you made these changes be equal??

Since we have been asked to find the value of $$2^{(1-x)^2}$$
It can be further simplified as $$2^{2 - 2x}$$ which is nothing but $$\frac{2^2}{2^{2x}}$$

Hope this helps you!
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Joined: 07 Apr 2018
Posts: 89
Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ? [#permalink]

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06 May 2018, 05:58
Reni wrote:
If $$2^{4x} = 3,600$$, what is the value of $$(2^{(1-x)})^2$$ ?

(A) -1/15
(B) 1/15
(C) 3/10
(D) -3/10
(E) 1

To find: (2^(1-x))^2
=> 2^(2-2x) = 4/ 2^2x ........................................(1)

Given: 2^4x= 3600 = 2^4 * 3^2 * 5^2
taking square root, we get
2^2x= 2^2 * 3 * 5..................................................(2)

Putting the value of 2^2x from eqn(2) to eqn(1), we get

4/2^2x = 4/(4 * 3 * 5) = 1/15

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Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ?   [#permalink] 06 May 2018, 05:58
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