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What is the value of x2 – 1 when 9x+ 1 = 27x– 1 ?
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15 May 2018, 16:07
15 May 2018, 16:07
Expert Reply
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Question Stats:
86% (01:06) correct
13% (00:55) wrong based on 15 sessions
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Re: What is the value of x2 – 1 when 9x+ 1 = 27x– 1 ?
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16 May 2018, 21:11
16 May 2018, 21:11
1
At first this problem looks complicated, but let's simplify this a little.
When we see \(9^{x+ 1} = 27^{x - 1}\), we can change it like so. \(3^2(x+ 1) = 3^3(x - 1)\).
How? \(3^2 = 9\) and \(3^3=27\).
From there, you can focus exclusively on the exponents. \(2(x+1)=3(x-1)\)
\(2x + 2 = 3x - 3\)
\(5 = x\)
Plus in 5 for the original equation.
\(x^2-1=?\)
\(5^2-1=24\)
The value of \(x^2 - 1\) when \(9^{x+ 1} = 27^{x - 1}\) is 24.
When we see \(9^{x+ 1} = 27^{x - 1}\), we can change it like so. \(3^2(x+ 1) = 3^3(x - 1)\).
How? \(3^2 = 9\) and \(3^3=27\).
From there, you can focus exclusively on the exponents. \(2(x+1)=3(x-1)\)
\(2x + 2 = 3x - 3\)
\(5 = x\)
Plus in 5 for the original equation.
\(x^2-1=?\)
\(5^2-1=24\)
The value of \(x^2 - 1\) when \(9^{x+ 1} = 27^{x - 1}\) is 24.
gmatclubot
Re: What is the value of x2 – 1 when 9x+ 1 = 27x– 1 ? [#permalink]
16 May 2018, 21:11
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