How do you solve #log x=-2#?

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Answer 1 · 2016-02-08T01:01:12

Since there is no base noted in subscript to the right of the log, this logarithm is in base 10. We can convert directly to exponential form.

#logx = -2#

#x = 10^-2#

#x = 1/(10^2)#

#x = 1/100#

The answer is #x = 1/100#

Practice exercises:

Solve the following logarithmic equations. Round solutions to the nearest hundredth whenever necessary.

a) #log(x + 1) = 1#

b) #log_7(x - 2) = 2#

Good luck!