How do you solve the equation ${log}_{5} 64 - {log}_{5} (\frac{8}{3}) + {log}_{5} 2 = {log}_{5} (4 p)$ $log_5 64-log_5 (8/3)+log_5 2=log_5 (4p)$ ?

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Answer 1 · 2016-11-15T13:50:43

$p = 12$ $p=12$

We have:

${log}_{5} (64) - {log}_{5} (\frac{8}{3}) + {log}_{5} (2) = {log}_{5} (4 p)$ $log_5(64)-log_5(8/3)+log_5(2)=log_5(4p)$

Let's first combine the left side:

${log}_{5} (\frac{64 \times 2}{\frac{8}{3}}) = {log}_{5} (4 p)$ $log_5((64xx2)/(8/3))=log_5(4p)$

${log}_{5} (\frac{64 \times 2 \times 3}{8}) = {log}_{5} (4 p)$ $log_5((64xx2xx3)/(8))=log_5(4p)$

${log}_{5} (8 \times 2 \times 3) = {log}_{5} (4 p)$ $log_5(8xx2xx3)=log_5(4p)$

${log}_{5} (48) = {log}_{5} (4 p)$ $log_5(48)=log_5(4p)$

We can now take the inverse function (effectively getting rid of the logs):

$48 = 4 p$ $48=4p$

$p = 12$ $p=12$

How do you solve the equation log564−log5(83)+log52=log5(4p)log_5 64-log_5 (8/3)+log_5 2=log_5 (4p)?