How do you solve ${log}_{2} (2 x - 3) = {log}_{2} (x + 4)$ $log_2(2x-3)=log_2(x+4)$ ?

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Answer 1 · 2017-02-13T15:52:14

$x = 7$ $x = 7$ .

If ${log}_{a} b = {log}_{a} c$ $log_a b = log_a c$ , then $b = c$ $b = c$ .

$2 x - 3 = x + 4$ $2x - 3= x +4$

$2 x - x = 4 + 3$ $2x - x = 4 + 3$

$x = 7$ $x = 7$

Hopefully this helps!

How do you solve log_2(2x-3)=log_2(x+4)log2(2x−3)=log2(x+4)log_2(2x-3)=log_2(x+4)?