If #logx = log1/2#, what is #x#?

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Answer 1 · 2018-04-09T02:51:40

#x = 1#

#log(x) = log(1)/2#

#log(x) = 0#

If this is #log equiv ln#:

#ln(x) = 0#

#x = e^0 = 1#

If this is #log equiv log_b#:

#log_b(x) = 0#

#x = b^0 = 1#

So for any logarithm, the solution is #x = 1#.

Answer 2 · 2018-04-09T02:53:24

#logx = log1/2#

#log1 = 0# since #10^0 = 1#.

Thus,
#logx = 0/2#
#log x = 0#
#10^(logx) = 10^0#
#x = 1#