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

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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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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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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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)

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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\(a^m^n=a^{(m^n)}\) 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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    Units digits, exponents, remainders problems
    Tough and tricky DS exponents and roots questions with detailed solutions
    Tough and tricky PS exponents and roots questions with detailed solutions
    Exponents DS Questions
    Exponents PS Questions
    Roots DS Questions
    Roots PS Questions

Check below for more:
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Ultimate GMAT Quantitative Megathread
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