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

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

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Updated on: 25 May 2017, 13:08
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Difficulty:

55% (hard)

Question Stats:

71% (02:28) correct 29% (03:16) wrong based on 129 sessions

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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

Originally posted by niteshwaghray on 25 May 2017, 13:00.
Last edited by Bunuel on 25 May 2017, 13:08, edited 1 time in total.
Renamed the topic and edited the question.
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Posts: 51229
Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) =  [#permalink]

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25 May 2017, 13:16
2
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, $$(x)^{(y+1)}=4^{1+1}=16$$.

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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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26 May 2017, 08:06
2
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, $$(x)^{(y+1)}=4^{1+1}=16$$.

Is it correct in the first step to assume that X = 4 just noticing that 1 doesn't appear in the possible answers? Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 51229
Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) =  [#permalink]

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26 May 2017, 08:52
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, $$(x)^{(y+1)}=4^{1+1}=16$$.

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]

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31 May 2017, 18:14
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.

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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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01 Jun 2017, 01:33
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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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) =  [#permalink]

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05 Oct 2018, 20:19
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Re: If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = &nbs [#permalink] 05 Oct 2018, 20:19
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