How do you solve ${log}_{x} 9 = - 2$ $log_x 9 = -2$ ?

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Answer 1 · 2016-03-07T03:25:22

You must change to exponential form: ${log}_{a} n = x \to a^{x} = n$ $log_an = x -> a^x = n$

$x^{- 2} = 9$ $x^-2 = 9$

Make the exponent positive by using the rule $x^{- n} = \frac{1}{x^{n}}$ $x^-n = 1/x^n$

$\frac{1}{x^{2}} = 9$ $1/x^2 = 9$

$1 = 9 x^{2}$ $1 = 9x^2$

$\frac{1}{9} = x^{2}$ $1/9 = x^2$

$\sqrt{\frac{1}{9}} = x$ $sqrt(1/9) = x$

$\frac{1}{3} = x$ $1/3 = x$ (a negative answer is not possible since the log of a negative number is undefined)

Practice exercises:

Challenge problem

What is the value of x in ${log}_{x} 4 = x$ $log_x4 = x$ ?

How do you solve logx9=−2log_x 9 = -2?