Question

How do you write ${\log _9}27 = \dfrac{3}{2}$ in exponential form?

Answer 1

Hint: To change from logarithmic form to exponential form, identify the base of the logarithmic equation and move the base to the other side of the equal sign. Moving the base will make the current number or variable into the exponent. Do not move anything but the base, the other numbers or variables will not change sides and the word “log” will be dropped. Identifying and moving the base is the key to changing from logarithmic form into exponential form.

Complete step by step solution:
A logarithm is an exponent which indicates to what power a base must be raised to produce a given number.
 $y = {b^x}$ exponential form,
$x = {\ln _b}y$ logarithmic function, where $x$ is the logarithm of $y$ to the base $b$, and${\log _b}y$ is the power to which we have to raise $b$ to get $y$, we are expressing $x$ in terms of $y$.
Given function is ${\log _9}27 = \dfrac{3}{2}$,
Here base is 9, now rewriting the left hand side of the equation using logarithmic identity, ${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$, we get,
$ \Rightarrow \dfrac{{\log 27}}{{\log 9}} = \dfrac{3}{2}$,
Now cross multiplying we get,
$ \Rightarrow 2\log 27 = 3\log 9$,
Now using logarithmic identity ${\log _a}{x^n} = n{\log _a}x$, we get,
$ \Rightarrow \log {27^2} = \log {9^3}$,
Now taking out the logarithms we get,
$ \Rightarrow {27^2} = {9^3}$,
Now taking square root on both sides we get,
$ \Rightarrow \sqrt {{{27}^2}} = \sqrt {{9^3}} $,
Now simplifying we get,
$ \Rightarrow 27 = {\left( {{9^3}} \right)^{\dfrac{1}{2}}}$,
Now simplifying we get,
$ \Rightarrow 27 = {9^{\dfrac{3}{2}}}$,
So, the exponential form is ${9^{\dfrac{3}{2}}}$.

Final Answer:
$\therefore $The exponential form of the given logarithmic function ${\log _9}27 = \dfrac{3}{2}$ will be equal to ${9^{\dfrac{3}{2}}}$.

Note:
A logarithm is a mathematical operation that determines how many times a certain number, called the base, is multiplied by itself to reach another number, we know that logarithm is the power to which a number must be raised in order to get some other number, and the base unit is the number being raised to a power. In these types of questions, we use logarithmic properties and formulas, and some of useful formulas are:
 ${\log _a}xy = {\log _a}x + {\log _a}y$,
$\log x - \log y = \log \left( {\dfrac{x}{y}} \right)$
${\log _a}{x^n} = n{\log _a}x$,
${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$,
${\log _{\dfrac{1}{a}}}b = - {\log _a}b$,
${\log _a}a = 1$,
${\log _{{a^x}}}b = \dfrac{1}{x}{\log _a}b$.