Question

Rewrite the expression as a single logarithm.
$\dfrac{1}{3}\ln x + 4\left[ {\ln (x - 3) - \dfrac{3}{8}\ln (x + 3)} \right]$
A. $\ln \left( {\dfrac{{8x(x - 3)}}{{9(x + 3)}}} \right)$
B. $\ln \left( {\dfrac{{\sqrt[3]{x}{{(x - 3)}^4}}}{{\sqrt {{{(x + 3)}^4}} }}} \right)$
C. $\ln \left( {\dfrac{{\sqrt[3]{x}{{(x - 3)}^4}}}{{\sqrt {{{(x + 3)}^3}} }}} \right)$
D. $\ln \left( {\dfrac{{\sqrt[3]{x}{{(x - 3)}^4}}}{{\sqrt[8]{{{{(x + 3)}^3}}}}}} \right)$

Answer 1

To solve this problem, we must learn all the laws of Logarithms first. The laws required to solve this problem, are the law of addition, the law of subtraction, and the law of power. If we convert all the coefficients to power, then it becomes very easy to use the law of subtraction and law of addition.

Formula Used:
Law of Addition: $\ln a + \ln b = \ln (ab)$
Law of Subtraction: $\ln a - \ln b = \ln \left( {\dfrac{a}{b}} \right)$
Law of Power: $b\ln a = \ln {a^b}$

Complete step-by-step answer:
Here, we assume that the base of all terms is the same because addition and subtraction are not possible for the terms with different log bases. The equation we are given here is
$\Rightarrow$ $\dfrac{1}{3}\ln x + 4\left[ {\ln (x - 3) - \dfrac{3}{8}\ln (x + 3)} \right]$
Now, we open the brackets and distribute the coefficient to both terms
$\Rightarrow \dfrac{1}{3}\ln x + 4\ln (x - 3) - 4 \times \dfrac{3}{8}\ln (x + 3)$
Focusing on the coefficient of the last term, by factoring we get
$\Rightarrow - 4 \times \dfrac{3}{{4 \times 2}} = - \dfrac{3}{2}$
Hence, the equation now is
$\Rightarrow \dfrac{1}{3}\ln x + 4\ln (x - 3) - \dfrac{3}{2}\ln (x + 3)$
Now, we need to apply the law of power. In the law of power, if the power is greater than or equal to $\;1$ , then it is written as it is.
But if the power is less than $\;1$ , then it is written as a root. For example, if the power is $\dfrac{1}{6}$, then it is written in roots as $\sqrt[6]{{}}$ .
Let’s consider the terms one-by-one to avoid confusion
For $\dfrac{1}{3}\ln x$ , the co-efficient is $\dfrac{1}{3}$ which converts to power as shown
$\Rightarrow \ln {x^{\tfrac{1}{3}}}$
As the power is less than $\;1$ , the power converts to root as shown
$\Rightarrow \ln \sqrt[3]{x}$
Now, let's consider the second term $4\ln (x - 3)$ . Here the co-efficient gets converted to power as shown
$\Rightarrow \ln {(x - 3)^4}$
As power is greater than $\;1$ , it remains as it is.
For the last term $\dfrac{3}{2}\ln (x + 3)$ , the co-efficient gets converted to power as shown
$\Rightarrow \ln {(x + 3)^{\tfrac{3}{2}}}$
Here, even though the number is greater than $\;1$ , we do not keep it as it is.
As the power is not an integer, hence we need to split the power as shown
$\Rightarrow \dfrac{3}{2} = 3 \times \dfrac{1}{2}$
Now, the part greater than $\;1$ is kept as it is, while the part less than $\;1$ is changed to roots as shown
$\Rightarrow \ln \sqrt {{{(x + 3)}^3}}$
Combining all the terms,
$\Rightarrow \ln \sqrt[3]{x} + \ln {(x - 3)^4} - \ln \sqrt {{{(x + 3)}^3}}$
Applying the law of addition, for the first two terms we get
$\Rightarrow \ln \sqrt[3]{x} + \ln {(x - 3)^4} = \ln \sqrt[3]{x}{(x - 3)^4}$
Now, substituting we get,
$\Rightarrow \ln \sqrt[3]{x}{(x - 3)^4} - \ln \sqrt {{{(x + 3)}^3}}$
Applying the law of subtraction, we get
$\Rightarrow \ln \sqrt[3]{x}{(x - 3)^4} - \ln \sqrt {{{(x + 3)}^3}} = \ln \left( {\dfrac{{\sqrt[3]{x}{{(x - 3)}^4}}}{{\sqrt {{{(x + 3)}^3}} }}} \right)$

Hence, converting the logarithm to a single term, we get Option C.

Note:
Here, the fact to remember is that the base of the logarithm must be checked before starting the problem. If the base is different, none of the laws of the logarithm can be applied. Here, we applied the law of addition first. But, if we apply the law of subtraction first, then also the answer remains the same.