How do you solve ${log}_{7} x = - 1$ $log_7 x= -1$ ?

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Answer 1 · 2016-05-11T11:50:55

$x = 7^{- 1} = \frac{1}{7}$ $x = 7^(-1) = 1/7$

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

If $x = b^{y}$ $" " x=b^y " "$ then ${log}_{b} (x) = y$ $" " log_b(x) = y$

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

$b^{{log}_{b} (x)} = b^{y} = x$ $b^(log_b(x)) = b^y= x$

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

${log}_{7} x = - 1$ $log_7x = -1$

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

$7^{{log}_{7} x} = 7^{- 1}$ $7^(log_7x) = 7^(-1)$

Then simplifying both sides using our expression from above:

$x = 7^{- 1} = \frac{1}{7}$ $x = 7^(-1) = 1/7$

How do you solve log7x=−1log_7 x= -1?