If $3^{n} = 27$ $3^n=27$ , what is the value of $4^{n - 1}$ $4^(n-1)$ ?

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Answer 1 · 2016-12-09T12:04:44

16

$3^{n} = 27$ $3^n=27$

Notice that $27 = 3^{3}$ $27 = 3^3$

Hence: $3^{n} = 3^{3}$ $3^n=3^3$

Equating exponents: $\to n = 3$ $-> n=3$

$:. 4^(n-1) = 4^(3-1)$

$=4^2 = 16$

If 3^n=273n=273^n=27, what is the value of 4^(n-1)4n−14^(n-1)?