How do you solve ${log}_{4} a + {log}_{4} 5 = 3$ $log_4 a + log_4 5 = 3$ ?

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How do you solve ${log}_{4} a + {log}_{4} 5 = 3$ $log_4 a + log_4 5 = 3$ ?

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Answer 1 · 2016-02-28T00:42:03

$a = 12.8$ $a=12.8$

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[1] $XXX {log}_{b} p + {log}_{b} q = {log}_{b} (p q)$ $color(white)("XXX")log_bp+log_bq = log_b(pq)$
[2] $XXX {log}_{b} m = k$ $color(white)("XXX")log_b m = k$ $XXXX$ $color(white)("XXXX")$ means $XXX b^{k} = m$ $color(white)("XXX")b^k=m$

${log}_{4} a + {log}_{4} 5 = {log}_{4} 5 a = 3$ $log_4 a + log_4 5 = log_4 5a = 3$ from [1] and given equality

$4^{3} = 5 a$ $4^3=5a$ from [2]

$\to 5 a = 64$ $rarr 5a = 64$

$\to a = \frac{64}{5} = 12.8$ $rarr a=64/5 = 12.8$

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How do you solve ${log}_{4} a + {log}_{4} 5 = 3$ $log_4 a + log_4 5 = 3$ ?

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