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If x and y are distinct positive integers, such that (2^x)(16^(x-1)) =

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niteshwaghray
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If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   Updated on: 25 May 2017, 14:08
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If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64
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C

Originally posted by niteshwaghray on 25 May 2017, 14:00.
Last edited by Bunuel on 25 May 2017, 14:08, edited 1 time in total.
Renamed the topic and edited the question.
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Bunuel
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   25 May 2017, 14:16
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niteshwaghray wrote:
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64


Step 1:



\((2^x)(16^{x−1})=2^{x^2}\);

\((2^x)(2^{4(x−1)})=2^{x^2}\);

\(2^{x+4(x−1)})=2^{x^2}\);

\(x+4(x−1)=x^2\);

\(x=1\) or \(x=4\)

Step 2:



\((3^y)(3^x)=3^{x^2−4x+5}\);

\((3^{x+y}=3^{x^2−4x+5}\);

\(x+y=x^2−4x+5\);

\(x^2-5x+5=y\).

Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\).

Step 3:



Therefore, \((x)^{(y+1)}=4^{1+1}=16\).

Answer: C.
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   26 May 2017, 09:06
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Bunuel wrote:
niteshwaghray wrote:
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64


Step 1:



\((2^x)(16^{x−1})=2^{x^2}\);

\((2^x)(2^{4(x−1)})=2^{x^2}\);

\(2^{x+4(x−1)})=2^{x^2}\);

\(x+4(x−1)=x^2\);

\(x=1\) or \(x=4\)

Step 2:



\((3^y)(3^x)=3^{x^2−4x+5}\);

\((3^{x+y}=3^{x^2−4x+5}\);

\(x+y=x^2−4x+5\);

\(x^2-5x+5=y\).

Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\).

Step 3:



Therefore, \((x)^{(y+1)}=4^{1+1}=16\).

Answer: C.


Is it correct in the first step to assume that X = 4 just noticing that 1 doesn't appear in the possible answers? Thanks!
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If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   10 Feb 2019, 20:09
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jfranciscocuencag wrote:
niteshwaghray wrote:
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64


Good night everyone!

Could someone please explain to me why \(2^{x^2}\) does not mean the following?

(2^m)^n = 2^mn

Isn't \(2^{x^2}\) the same as (2^x)^2?

Thank you so much in advance!8


Hey jfranciscocuencag

(2^m)^n = 2^mn , The presence of bracket makes all the difference, As per PEMDAS, we solve the parenthesis first

But if you see here, there is no parenthesis, because of that i will first solve the exponent and not the expression \(2^{x^2}\)

Have a look here as well
https://www.mathsisfun.com/operation-order-pemdas.html
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   26 May 2017, 09:52
Expert Reply
MvArrow wrote:
Bunuel wrote:
niteshwaghray wrote:
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64


Step 1:



\((2^x)(16^{x−1})=2^{x^2}\);

\((2^x)(2^{4(x−1)})=2^{x^2}\);

\(2^{x+4(x−1)})=2^{x^2}\);

\(x+4(x−1)=x^2\);

\(x=1\) or \(x=4\)

Step 2:



\((3^y)(3^x)=3^{x^2−4x+5}\);

\((3^{x+y}=3^{x^2−4x+5}\);

\(x+y=x^2−4x+5\);

\(x^2-5x+5=y\).

Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\).

Step 3:



Therefore, \((x)^{(y+1)}=4^{1+1}=16\).

Answer: C.


Is it correct in the first step to assume that X = 4 just noticing that 1 doesn't appear in the possible answers? Thanks!


Good observation. If x were 1, then the answer would be 1 because 1^(any number) = 1. So, yes we could say after Step 1, that x = 4.
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   31 May 2017, 19:14
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niteshwaghray wrote:
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64


Let’s simplify the first equation:

(2^x)(16^x-1) = 2^(x^2)

(2^x)(2^4)^(x-1) = 2^(x^2)

(2^x)(2^(4x-4)) = 2^(x^2)

2^(5x - 4) = 2^(x^2)

Because the bases are each 2, we can equate just the exponents:

5x - 4 = x^2

x^2 - 5x + 4 = 0

(x - 4)(x - 1) = 0

x = 4 or x = 1

Now let’s simplify the second equation:

(3^y)(3^x) = 3^(x^2 - 4x + 5)

3^(y + x) = 3^(x^2 - 4x + 5)

y + x = x^2 - 4x + 5

y = x^2 - 5x + 5

Recall from our earlier work that x^2 - 5x + 4 = 0; thus, y = x^2 - 5x + 5 = (x^2 - 5x + 4) + 1 = 0 + 1 = 1. Furthermore, recall that x = 4 or x = 1 and we are given that x and y are distinct integers, so x must be equal to 4. Thus, x^(y + 1) = 4^(1 + 1) = 4^2 = 16.

Answer: C
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   01 Jun 2017, 02:33
Since \((2^x)(16^{x−1})=2^{x^2}\)
We can clearly say that \(x+4x-4 = x^2\) because \(16 = 2^4\)
Therefore, \(x^2 = 5x-4\)

\((3^y)(3^x)=3^{x^2−4x+5}\)
Hence,\(y+x = {x^2−4x+5}\)
Substituting value of x^2 in this equation
y + x = 5x - 4 -4x + 5
y + x = x + 1 or y=1.

Substituting y=1, \((x)^{(y+1)}\) must be a perfect square.
From equation \(x^2 = 5x-4\) we can find out the value of x to be 1 or 4.
Only 4^2 is an option given in the answer options(Option C)
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If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   10 Feb 2019, 19:53
niteshwaghray wrote:
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64


Good night everyone!

Could someone please explain to me why \(2^{x^2}\) does not mean the following?

(2^m)^n = 2^mn

Isn't \(2^{x^2}\) the same as (2^x)^2?

Thank you so much in advance!8
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Bunuel
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   10 Feb 2019, 21:55
Expert Reply
jfranciscocuencag wrote:
niteshwaghray wrote:
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64


Good night everyone!

Could someone please explain to me why \(2^{x^2}\) does not mean the following?

(2^m)^n = 2^mn

Isn't \(2^{x^2}\) the same as (2^x)^2?

Thank you so much in advance!8


\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down). Thus, \(2^{x^2}=2^{(x^2)}\) and not \((2^x)^2=2^{2x}\)

On the other hand, \((a^m)^n=a^{mn}\).

8. Exponents and Roots of Numbers



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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink] New post   30 Nov 2021, 09:41
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = [#permalink]
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