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Suppose $81^x = 64$. What is $27^{x+1}$?

 Jul 11, 2020
 #1
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81^x = 64 ==> \(x = \log_{81} 64\)

 

Then \(27^{x + 1} = 27^{\log_{81} 64 + 1} = 243\)

 Jul 11, 2020

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