How do you solve ${log}_{4} z + {log}_{4} 8 = {log}_{4} 24$ $log_4 z + log_4 8 = log_4 24$ ?

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Answer 1 · 2018-07-29T06:43:52

$z = 3$ $color(red)(z = 3$

$log a + log b = log (a b), log rule$ $log a + log b = log (ab), " Log rule"$

${log}_{4} z + {log}_{4} 8 = {log}_{4} 24$ $log_4 z + log_4 8 = log_4 24$

${log}_{4} (8 z) = {log}_{4} 24$ $log_4(8 z) = log_4 24$

$:. 8 z = 24, " as base same"$

$color(red)(z = 24/8 = 3$

How do you solve log_4 z + log_4 8 = log_4 24log4z+log48=log424log_4 z + log_4 8 = log_4 24?