How do you write #log_x(64)=3# in exponential form?

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Answer 1 · 2018-05-09T12:37:49

#x^3=64#

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

#•color(white)(x)log_b x=nhArrx=b^n#

#"here "x=64,b=x" and "n=3#

#rArrlog_x 64=3rArrx^3=64#

Answer 2 · 2018-05-09T13:12:13

#64=4^3#

#log_x (64) = 3#

#log_x (4^3 )=3#

#3log_x 4 = 3 -> x=4#

Hence, we can express the identity in exponential form as: #64=4^3#