How do you solve ${log}_{4} 5 - {log}_{4} (- 4 x) = 1$ $log_4 5 - log_4 (-4x)=1$ ?

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Answer 1 · 2016-10-20T16:38:16

$x = - \frac{5}{16}$ $x = - 5/16$

${log}_{4} 5 - {log}_{4} (- 4 x) = 1$ $log_4 5 - log_4 (-4x) = 1$

${log}_{4} 5 - {log}_{4} (- 4 x) = {log}_{4} 4$ $log_4 5 - log_4 (-4x) = log_4 4$

${log}_{4} (\frac{5}{- 4 x}) = {log}_{4} 4$ $log_4 (5/(-4x)) = log_4 4$

$(\frac{5}{- 4 x}) = 4$ $(5/(-4x)) = 4$

$x = - \frac{5}{16}$ $x = -5/16$

How do you solve log45−log4(−4x)=1log_4 5 - log_4 (-4x)=1?