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 $\log x$ .