Question

Solve for x: $ \dfrac{1}{3}\ln x + \ln 2 - \ln 3 = 3 $ .

Answer 1

Hint: Log to the base of e is the natural logarithm, which is denoted by ln. e is also called as the natural number, Euler’s number, natural exponent which is a mathematical constant, an irrational and transcendental number which is equal to $ 2.71828 $ approximately. So, $ \ln \left( x \right) = {\log _e}\left( x \right) $ .

Formula used:
(1) $ {e^{ - x}} = \dfrac{1}{{{e^x}}} $
(2) $ n\ln \left( a \right) = \ln \left( {{a^n}} \right) $
 (3) $ {e^{\ln \left( a \right)}} = a $ if $ a > 0 $

Complete step by step solution:
In this problem, we have to solve for x. The given equation is, $ \dfrac{1}{3}\ln x + \ln 2 - \ln 3 = 3 $
Firstly, we will rearrange the above equation,
 $
   \Rightarrow 3 + \ln 3 - \ln 2 = \dfrac{1}{3}\ln x \\
   \Rightarrow 3\left( {3 + \ln 3 - \ln 2} \right) = \ln x \;
  $
Now, we will apply the exponential function in the above rearranging equation, on applying, we get,
 $ \Rightarrow x = {e^{3 \times \left( {3 + \ln \left( 3 \right) - \ln \left( 2 \right)} \right)}} $
On further simplifying, we get,
 $ \Rightarrow x = {e^{9 + 3\ln \left( 3 \right) - 3\ln \left( 2 \right)}} $
Here, we have used the formulas to solve this, from formula (2) $ n\ln \left( a \right) = \ln \left( {{a^n}} \right) $ , $ 3\ln \left( 3 \right) $ becomes $ \ln \left( {{3^3}} \right) $ and from formula (3) $ {e^{\ln \left( a \right)}} = a $ if $ a > 0 $ , $ {e^{\ln \left( {{3^3}} \right)}} = {3^3} $ which becomes $ 27 $ . Similarly, from formula (2) $ n\ln \left( a \right) = \ln \left( {{a^n}} \right) $ , $ 3\ln \left( 2 \right) $ becomes $ \ln \left( {{2^3}} \right) $ , from formula (1) $ {e^{ - x}} = \dfrac{1}{{{e^x}}} $ , $ {e^{ - \ln \left( {{2^3}} \right)}} $ becomes $ \dfrac{1}{{{e^{\ln \left( {{2^3}} \right)}}}} $ and from formula (3) $ {e^{\ln \left( a \right)}} = a $ if $ a > 0 $ , $ \dfrac{1}{{{e^{\ln \left( {{2^3}} \right)}}}} $ becomes $ \dfrac{1}{{{2^3}}} $ which is equal to $ \dfrac{1}{8} $ .
 By using the formulas given above and on further solving, we get,
 $ \Rightarrow x = {e^9} \times 27 \times \dfrac{1}{8} $
Which results into,
 $ \Rightarrow \dfrac{{27}}{8}{e^9} $ .
Hence, the value of x in the equation given in the question is $ \dfrac{{27}}{8}{e^9} $ .
So, the correct answer is “ $ \dfrac{{27}}{8}{e^9} $ ”.

Note: The form of the exponential function, which is a mathematical function, is $ f\left( x \right) = {a^x} $ , where a is a constant which is known as the base of the function and it should be greater than $ 0 $ and x is a variable. The log that is usually used in higher mathematics is the natural logarithm. The natural logarithm of x is written as $ \ln x,{\log _e}x $ and if the base e is implicit then we can simply write it as $ \log x $ .