How do you solve #log_3 root4(5)<=x#?

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Answer 1 · 2017-01-07T14:24:23

#x>= 0.366#

#log_3 root(4)5 <= x# can be written as

#x >= log_3 root(4)5#

or #x >= log_3 (5)^(1/4)#

or #x >= 1/4log_3 5#

or #x >= 1/4 log5/log3#

or #x >= 1/4xx0.6990/0.4771#

or #x >= 1/4xx1.465#

or #x>= 0.366#