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

making the answer 1.

Why is this not the correct way to approach?

Thanks in advance

\(2^{4x} = 2^4*3^2*5^2\). 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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If 2^4x = 3,600, what is the value of (2^1-x)^2 ?
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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:)

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

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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Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ?
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12 Oct 2018, 06:48

1

Top Contributor

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) = 3,600 Raise both sides to the power of 1/2 to get: [2^(4x)]^(1/2) = 3,600^(1/2) Simplify: 2^(2x) = 60

We want to find the value of: [2^(1−x)]² To simplify, apply Power of Power law to get: 2^(2 - 2x) This is equal to: (2^2)/[2^(2x)] Replace 2^(2x) with 60 to get: (2^2)/60 = 4/60 = 1/15

Re: If 2^4x = 3,600, what is the value of (2^1-x)^2 ?
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15 Oct 2019, 17:09

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