How do you simplify $64^{{log}_{4} (8 y)}$ $64^(log_4 (8y))$ ?

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Answer 1 · 2015-12-31T06:50:15

$512 y^{3}$ $512y^3$

$64 = 4^{3}$ $64=4^3$ , so the expression can be written as

$\Rightarrow {(4^{3})}^{{log}_{4} (8 y)}$ $=>(4^3)^(log_4(8y))$

$\Rightarrow 4^{3 {log}_{4} (8 y)}$ $=>4^(3log_4(8y))$

Rewrite using the rule: $a {log}_{b} (c) = {log}_{b} (c^{a})$ $alog_b(c)=log_b(c^a)$

$\Rightarrow 4^{{log}_{4} ({(8 y)}^{3})}$ $=>4^(log_4((8y)^3))$

The $4$ $4$ and the ${log}_{4}$ $log_4$ will cancel since exponentiation and logarithmic functions are inverses.

$a^{{log}_{a} (b)} = b$ $a^(log_a(b))=b$

$\Rightarrow {(8 y)}^{3}$ $=>(8y)^3$

$\Rightarrow 512 y^{3}$ $=>512y^3$

How do you simplify 64log4(8y)64^(log_4 (8y))?