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} $