How do you solve ${log}_{2} (24) - {log}_{2} 3 = {log}_{5} (x)$ $log_2(24) - log_2 3 = log_5(x)$ ?

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How do you solve ${log}_{2} (24) - {log}_{2} 3 = {log}_{5} (x)$ $log_2(24) - log_2 3 = log_5(x)$ ?

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Answer 1 · 2015-12-21T17:54:32

$x = 125$ $x=125$

Explanation:

${log}_{2} (24) - {log}_{2} (3) = {log}_{2} (\frac{24}{3}) = {log}_{2} (8) = 3$ $log_2(24)-log_2(3)= log_2(24/3) = log_2(8) = 3$
$XXXXXXXXXXXXXXXXXXXXXXXX$ $color(white)("XXXXXXXXXXXXXXXXXXXXXXXX")$ since 2^3=8#

Therefore
$XXX {log}_{2} (24) - {log}_{2} (3) = {log}_{5} (x)$ $color(white)("XXX")log_2(24)-log_2(3)=log_5(x)$
$\to$ $rarr$
$XXX {log}_{5} (x) = 3$ $color(white)("XXX")log_5(x) = 3$
$\to$ $rarr$
$XXX x = 5^{3} = 125$ $color(white)("XXX")x = 5^3 =125$

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How do you solve ${log}_{2} (24) - {log}_{2} 3 = {log}_{5} (x)$ $log_2(24) - log_2 3 = log_5(x)$ ?

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How do you solve ${log}_{2} (24) - {log}_{2} 3 = {log}_{5} (x)$ $log_2(24) - log_2 3 = log_5(x)$ ?