How do you solve the equation $2 log 4 (x + 7) - log 4 (16) = 2$ $2 log4(x + 7)-log4(16) = 2$ ?

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Answer 1 · 2016-08-11T21:36:17

$x = 9$ $x= 9$

When working with a log equation, the term must be all logs or all numbers.
Change the RHS to a log term as well.

$2 {log}_{4} (x + 7) - {log}_{4} 16 = {log}_{4} 16 2 = {log}_{4} 16$ $color(red)(2)log_4 (x+7) - log_4 16 = log_4 16 " "2 = log_4 16$

${log}_{4} {(x + 7)}^{2} - {log}_{4} 16 = {log}_{4} 16$ $log_4 (x+7)^color(red)(2) - log_4 16 = log_4 16$

"If logs are being subtracted, the numbers are being divided"

${log}_{4} (\frac{{(x + 7)}^{2}}{16}) = {log}_{4} 16$ $log_4 ((x+7)^2/16)= log_4 16$

$\frac{{(x + 7)}^{2}}{16} = 16 if log A = log B, A = B$ $(x+7)^2/16 = 16" if log A = log B, A = B"$

${(x + 7)}^{2} = 256 cross multiply$ $(x+7)^2 = 256" cross multiply"$

$x + 7 = 16 only pos root is valid$ $x+7 = 16" only pos root is valid"$

$x = 9$ $x= 9$

How do you solve the equation 2 log4(x + 7)-log4(16) = 22log4(x+7)−log4(16)=22 log4(x + 7)-log4(16) = 2?