What is the value of \(x^2 - 1\) when \(9^{x+ 1} = 27^{x - 1}\) ?

A. 8
B. 12
C. 18
D. 24
E. 32
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\(9^{x+ 1} = 27^{x - 1}\)

\(3^{2(x+ 1)} = 3^{3(x - 1)}\)
\(2x+ 2 = 3x - 3\)
\(5 = x\)

Hence
\(x^2 - 1 = 5^2 - 1 = 24\)

Option (D) is our bet.

Best,
GLadi

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Regards,
Gladi



“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)

No it does not. You are mixing two things up. One is the equation given by the question stem. On solving that, as you can see in my solution above, you can come to the conclusion that x = 5.

Second is what the question is asking, it does not ask us the value of x but instead, it asks x^2 - 1.

Equating x^2 - 1 to a product of (2x + 2) and (3x - 3) is bizarre and there is no logical reason why one should do that. That is why I am assuming you mixed things up.

Hope it helps.

_________________

Regards,
Gladi



“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)

In order to solve for x, we need a common base for each expression in the equation. Note that 9 = 3^2 and 27 = 3^3. Simplifying, we have:

3^(2x + 2) = 3^(3x - 3)

2x + 2 = 3x - 3

5 = x

So x^2 - 1 = 24.

Answer: D
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