Question 1032859
You do not need e here.


Use the fact that *[tex \large \log_{b} x + \log_b {y} = \log_b {xy}]. So *[tex \large \log_2 2x + \log_2 x = \log_2 2x^2 = 5].


Since *[tex \large \log_2 32 = 5], we must have *[tex \large 2x^2 = 32 \Leftrightarrow x^2 = 16]. Assuming we only define log on the positive real numbers, the only solution is *[tex \large x = 4] (x = -4 introduces logs of negative numbers).