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

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Answer 1 · 2016-07-24T15:08:34

x=3

${log}_{2} (x + 5) - {log}_{2} (x - 2) = 3$ $log_2(x+5)-log_2(x-2)=3$

combine the log terms as follows

${log}_{2} (\frac{x + 5}{x - 2}) = 3$ $log_2 ((x+5)/(x-2))=3$

then by definition

$(\frac{x + 5}{x - 2}) = 2^{3} = 8$ $((x+5)/(x-2))=2^3 = 8$

then some algebra

$x + 5 = 8 x - 16$ $x+5= 8x-16$

$7 x = 21, x = 3$ $7x = 21, x=3$

How do you solve log2(x+5)−log2(x−2)=3log_2(x+5)-log_2(x-2)=3?