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Solve 64^x = 16^(x−1)

 Oct 16, 2021
 #1
avatar+390 
+1

Hello,

 

64^x = 16^(x−1),

 

We can write 64^x to and 16^(x−1) to:

 

(\({4}^{3x}={4}^{2x-2}\)),

 

(\(3x=2x-2\)),

 

(\(3x-2x=-2\)),

 

(\(x=-2\))

 

Straight

 Oct 16, 2021
 #3
avatar+115338 
+1

That is a nice method Straight :)

Melody  Oct 16, 2021
 #2
avatar+115338 
+1

\(64^x = 16^{(x−1)}\\ log_2 64^x = log_2 16^{(x−1)}\\ xlog_2 64 =(x-1) log_2 16\\ x*6 =(x-1) *4\\ 6x=4x-4\\ 2x=-4\\ x=-2\\~\\ check\\ LHS=64^{-2}=(2^6)^{-2}=2^{-12}\\ RHS=16^{-2-1}=16^{-3}=(2^4)^{-3}=2^{-12}=LHS \)

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 Oct 16, 2021

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