Question

How do you write $ - 2 = {\log _9}\dfrac{1}{{81}}$ in exponential form?

Answer 1

Hint: As we have to write the expression in exponential form, use the definition of the logarithm, which says that when the logarithm in the form ${\log _b}x = y$ is converted into the exponential form is equivalent to ${b^y} = x$. After that compare the give logarithm value with this form to get the desired result.

Complete step by step answer:
The given expression is $ - 2 = {\log _9}\dfrac{1}{{81}}$.
Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
There are two types of logarithms which is the common logarithmic function and the natural logarithmic function.
Common Logarithmic Function or Common logarithm is the logarithm with a base equal to 10. It is also known as the decimal logarithm because of its base. The common logarithm of x is denoted as \[\log x\]. For example, $\log 10 = 1$ (Since, ${10^2} = 100$).
The Natural Logarithmic Function is the logarithm with a base equal to the mathematical constant $e$. The value of \[e\] which is a mathematical constant is approximately equal to 2.7182818. The common logarithm of x is denoted as \[{\log _e}x\] or $\ln x$. For example, ${\log _e}100 = \ln 100$.
By the definition of the logarithm, when the logarithm in the form ${\log _b}x = y$ is converted into the exponential form is equivalent to ${b^y} = x$.
Now compare the expression with the definition form to get the value of $x,b$ and $y$.
So, $x = \dfrac{1}{{81}},b = 9,y = - 2$.
So, the exponential form will be,
$ \Rightarrow {9^{ - 2}} = \dfrac{1}{{81}}$

Hence, the exponential form of $ - 2 = {\log _9}\dfrac{1}{{81}}$ is equivalent to ${9^{ - 2}} = \dfrac{1}{{81}}$.

Note: The different rules of the log are,
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Product rule law,
$\log xy = \log x + \log y$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$