How do you solve $- 2 = {log}_{x} (\frac{1}{100})$ $-2=log_x (1/100)$ ?

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How do you solve $- 2 = {log}_{x} (\frac{1}{100})$ $-2=log_x (1/100)$ ?

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Answer 1 · 2015-12-21T07:02:44

$x = 10$ $x= 10$

Explanation:

$- 2 = {log}_{x} (\frac{1}{100})$ $-2 = log_x (1/100)$

There are many ways you can solve this problem. The best approach is to convert the given equation to exponent form.

Rule : ${log}_{b} (a) = k \Rightarrow a = k^{b}$ $log_b (a) = k => a= k^b$

Using the rule we can get

$x^{- 2} = \frac{1}{100}$ $x^-2 = 1/100$
$x^{- 2} = \frac{1}{10^{2}}$ $x^-2 =1/10^2$
$x^{- 2} = 10^{- 2}$ $x^-2 = 10^-2$ Rule $\frac{1}{a^{m}} = a^{- m}$ $1/a^m = a^-m$

$x = 10$ $x= 10$ The final answer.

Alternate approach

$- 2 = {log}_{x} (\frac{1}{100})$ $-2 = log_x (1/100)$
$- 2 = {log}_{x} (10^{- 2})$ $-2 = log_x (10^-2)$
$- 2 = - 2 {log}_{x} (10)$ $-2 = -2 log_x (10)$ Rule ${log (A)}^{n} = n log (A)$ $log(A)^n = n log(A)$
$1 = {log}_{x} (10)$ $1 = log_x (10)$
$x = 10$ $x =10$ because of the rule ${log}_{a} (a) = 1$ $log_a (a) = 1$ x has to be 10.

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Explanation:

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