\(2^{4x} = 3600\)

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

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

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

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

\(= \frac{1}{15}\)

Answer = B

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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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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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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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

Answer: B
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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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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

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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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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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

Correct Answer- B

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

Answer: B

RELATED VIDEO FROM OUR COURSE

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This is what i donot understand. How can (2^1-x)^2 = 2^2/2^2x?

Please do not skip any step of the simplyfications in between en explain to me as you might to a child or a golden retriever.

Thanks in Advance!

(2^1-x)^2 = 2^2/2^2x
so, (a^m)^n = a^(m*n) Right?
so (2^1-x)^2=2^((1-x)*2)=2^(2-2x)
now, a^(m-n)=(a^m)/(a^n) okay?
thus,
2^(2-2x) = 2^2/2^2x
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prathyushaR

Thanks a lot for your Explaination! Now I understand this question!

2^4x=3600
2^2^2x = 60^2
2^2x = 60 (A)

Now we need to find
2^(1-x)^2 can we written as
(2/2x)^2
4/2^2x
Put value of A
4/60
1/15
Answer is B

keep in mind what we need to find:\((2^{(1-x)})^2\)
this can also be written as \((2^{-(x-1)})^2\)

\(2^{4x}=3^2*2^4*5^2\)
\(2^x=3^{1/2}*2*5^{1/2}\)
\(2^{(x-1)}=3^{1/2}*5^{1/2}\)
\((2^{(1-x)})^2\) = 3*5

but we want to find \((2^{-(x-1)})^2\) hence answer is reciprocal of 15 = 1/15

2^(4x) = 3600
(2^(2x))^2 = 60^2
2^(2x) = 60

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

Answer is B