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What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?

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What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?  [#permalink]

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10 Dec 2018, 23:32
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Difficulty:

15% (low)

Question Stats:

78% (01:10) correct 22% (01:53) wrong based on 49 sessions

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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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Re: What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?  [#permalink]

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10 Dec 2018, 23:49
Bunuel wrote:
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

Given ,

$$9^{x+ 1} = 27^{x - 1}$$

$$3^{2x + 2} = 3^{3x -3}$$

2x + 2 = 3x - 3

x = 5.

$$x^2 - 1$$

= $$5^2 - 1$$
= 24.

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What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?  [#permalink]

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10 Dec 2018, 23:50
$$9^{x+ 1} = 27^{x - 1}$$

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

Hence
$$x^2 - 1 = 5^2 - 1 = 24$$

Option (D) is our bet.

Best,
Bunuel wrote:
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

_________________

Regards,

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

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Joined: 12 Sep 2017
Posts: 135
What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?  [#permalink]

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05 Feb 2019, 19:10
Hello!

What's wrong with solving it like the following?

$$x^2 - 1$$ = (x+1)(x-1)

(x+1) = 2x +2
(x-1) = 3x - 3

Why it does not reach the same answer?

$$x^2 - 1$$ = (2x +2) (3x - 3)

Shouldn't be the same?

Kind regards!
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Status: It always seems impossible until it's done.
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Re: What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?  [#permalink]

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05 Feb 2019, 22:16
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.
jfranciscocuencag wrote:
Hello!

What's wrong with solving it like the following?

$$x^2 - 1$$ = (x+1)(x-1)

(x+1) = 2x +2
(x-1) = 3x - 3

Why it does not reach the same answer?

$$x^2 - 1$$ = (2x +2) (3x - 3)

Shouldn't be the same?

Kind regards!

_________________

Regards,

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

Manager
Joined: 12 Sep 2017
Posts: 135
Re: What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?  [#permalink]

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06 Feb 2019, 08:23
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.
jfranciscocuencag wrote:
Hello!

What's wrong with solving it like the following?

$$x^2 - 1$$ = (x+1)(x-1)

(x+1) = 2x +2
(x-1) = 3x - 3

Why it does not reach the same answer?

$$x^2 - 1$$ = (2x +2) (3x - 3)

Shouldn't be the same?

Kind regards!

Thank you very much Gladiator59 !

Yes, now that I read it again it doe not make more sense to me, I guess I was tired.

Kind regards!
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Re: What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?  [#permalink]

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08 Feb 2019, 07:03
Bunuel wrote:
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

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.

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Re: What is the value of x^2 – 1 when 9^(x + 1) = 27^(x – 1) ?   [#permalink] 08 Feb 2019, 07:03
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