How do you solve # log _ { 4} x = 3#?

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Answer 1 · 2018-04-16T08:31:18

# x=64#.

By the definition of #log.# function,

#log_b a=m iff b^m=a, b gt 0, b!=1#.

#:. log_4 x=3 rArr 4^3=x=64#.

Answer 2 · 2018-04-16T08:37:48

#x=64#

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)log_b x=nhArrx=b^n#

#"here "b=4" and "n=3#

#rArrx=4^3=64#