How do you write #32=x^5# in Logarithm form?

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Answer 1 · 2016-08-28T20:04:29

#32=x^5 iff 5=log_x 32#.

Answer 2 · 2016-08-28T20:04:43

It can be written:

#log_x 32 = 5 " "# or #" " log_32 x = 1/5#

Given:

#32 = x^5#

If we take logs base #x# we find:

#log_x 32 = log_x x^5 = 5#

Alternatively, if we take logs base #32# we find:

#1 = log_32 32 = log_32 x^5 = 5 log_32 x#

Hence:

#log_32 x = 1/5#