How do you solve #−log(x) = 5.4#?

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Answer 1 · 2018-06-22T06:24:12

#x=e^(-27/10)=x#

After5 the Definition

#log_a(b)=n# if #a^n=b#
we get

#log(x)=-27/5#
therefore

#x=e^(-27/5)#

Answer 2 · 2018-06-22T06:50:36

#~~3.98*10^-6#

Given: #-log(x)=5.4#

#=>log(x)=-5.4#

Since there is no base, the assumed base used is base #10#.

So, we get:

#log_10(x)=-5.4#

#x=10^-5.4# (through logarithm definitions)

#~~3.98*10^-6#