How do you solve #log_10x=-1#?

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Answer 1 · 2017-07-30T08:27:19

#x = frac(1)(10)#

We have: #log_(10)(x) = - 1#

A logarithmic number can be expressed in the following way:

#" " " " " " " " " " " " " " log_(a)(b) = y <=> a^(y) = b#

Using this method, we can isolate #x#, and hence solve the equation.

So let's do it:

#Rightarrow log_(10)(x) = - 1 <=> 10^(- 1) = x#

#Rightarrow log_(10)(x) = - 1 <=> frac(1)(10) = x#

#Rightarrow x = frac(1)(10)#

Therefore, the solution of the equation is #x = frac(1)(10)#.