How do you solve $2 {log}_{6} (4 x) = 0$ $2log_6(4x)=0$ ?

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Answer 1 · 2016-07-15T21:23:43

$x = \frac{1}{4}$ $x = 1/4$

By definition if ${log}_{a} b = c, then a^{c} = b$ $log_a b= c, " then " a^c = b$

$2 {log}_{6} (4 x) = 0$ $2log_6 (4x) =0$

$\frac{2 {log}_{6} (4 x)}{2} = \frac{0}{2}$ $(2log_6 (4x))/2 =0/2$

${log}_{6} 4 x = 0 \Rightarrow 6^{0} = 4 x$ $log_6 4x = 0" "rArr 6^0 = 4x$

$6^{0} = 4 x$ $6^0 = 4x$

$1 = 4 x$ $1 = 4x$

$x = \frac{1}{4}$ $x = 1/4$

How do you solve 2log_6(4x)=02log6(4x)=02log_6(4x)=0?