How do you solve log_7 x= -1log7x=1?

1 Answer
May 11, 2016

x = 7^(-1) = 1/7x=71=17

Explanation:

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

If " " x=b^y " " x=by then " " log_b(x) = y logb(x)=y

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

b^(log_b(x)) = b^y= xblogb(x)=by=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 = -1log7x=1

Raising both sides of the equation to the power of the base, 77, we get

7^(log_7x) = 7^(-1)7log7x=71

Then simplifying both sides using our expression from above:

x = 7^(-1) = 1/7x=71=17