How do you solve #log_4x=3#?
1 Answer
Jul 9, 2016
x = 64
Explanation:
#color(orange)"Reminder"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(log_b x=ahArrx=b^a)color(white)(a/a)|)))# here a = 3 and b = 4
#rArrlog_4 x=3rArrx=4^3=64#
