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:

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

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