How do you solve #log_7 x= -1#?

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Answer 1 · 2016-05-11T11:50:55

#x = 7^(-1) = 1/7#

The log function is asking us what power of the base gives us the argument of the #log#, in equation form:

If #" " x=b^y " " # then #" " log_b(x) = y#

The inverse of a #log# function is a power of the base, i.e.

#b^(log_b(x)) = b^y= x#

The way we use this is to apply the "inverse" to both sides of an equation. Starting with the equation from our question

#log_7x = -1#

Raising both sides of the equation to the power of the base, #7#, we get

#7^(log_7x) = 7^(-1)#

Then simplifying both sides using our expression from above:

#x = 7^(-1) = 1/7#