Question

How do you solve $\log \left( {\dfrac{1}{{{{10}^x}}}} \right)$?

Answer 1

Hint: In order to solve the above logarithmic function, use the property of logarithm that Fraction inside the logarithm can be expressed as the subtraction of log of numerator and log of denominator and property $\log {m^n} = n\log m$, and then use put the value of $\log (10) = 1$, $\log (1) = 0$, to obtain your required result.

Formula used:
${\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n)$
${\log _b}(m) + {\log _b}(n) = {\log _b}(mn)$
$n\log m = \log {m^n}$

Complete step by step answer:
We are given a logarithmic function $\log \left( {\dfrac{1}{{{{10}^x}}}} \right)$.
In order to simplify the above logarithmic function, we will take help from some of the logarithmic properties.
Fraction inside the logarithm can be expressed as the subtraction of log of numerator and log of denominator as ${\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n)$
Using this property or function becomes,
$ = \log (1) - \log ({10^x})$
Since value of $\log (1) = 0$ and applying property $\log {m^n} = n\log m$ in the second term, we get
$
   = 0 - x\log ({10^x}) \\
   = - x\log (10) \\
 $
Value of $\log (10) = 1$
$
   = - x(1) \\
   = - x \\
 $
Therefore, the simplification of logarithmic function $\log \left( {\dfrac{1}{{{{10}^x}}}} \right)$ is equal to $ - x$.

Note: 1. Value of constant ‘e’ is equal to $2.71828$.
2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number, we actually undo an exponentiation.
3. Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values.
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values.
${\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$