Question

How do you solve for ${\log _9}27 = x$ ?

Answer 1

Hint : In order to determine the value of the above question use the property of logarithm that ${\log _b}a$ is equivalent to $= \dfrac{{\ln a}}{{\ln b}}$ and consider another property $\log {m^n} = n\log m$ to get your desired answer accordingly.

Complete step-by-step answer :
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
First we are using identity ${\log _b}a = \dfrac{{\ln a}}{{\ln b}}$
$\Rightarrow {\log _9}27 = x \\\ \Rightarrow \dfrac{{\ln 27}}{{\ln 9}} = x \;$
Factorising $27$ ,so $27$ can be written as ${3^3}$ and similarly $9$ can be written as ${3^2}$
$\Rightarrow \dfrac{{\ln {3^3}}}{{\ln {3^2}}} = x$
Using property of logarithm that $\log {m^n} = n\log m$
$\Rightarrow \dfrac{{3\ln 3}}{{2\ln 3}} = x$
Cancelling $\ln 3$ from numerator and denominator
$\Rightarrow \dfrac{{3\operatorname{l} {n}3}}{{2\operatorname{l} {n}3}} = x \\\ \Rightarrow \dfrac{3}{2} = x \\\ \Rightarrow x = \dfrac{3}{2} \;$
Therefore, solution to expression ${\log _9}27 = x$ is equal to $x = \dfrac{3}{2}$
So, the correct answer is “ $x = \dfrac{3}{2}$ ”.

Note : 1. Value of the 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 .
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$