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How do you simplify $\ln 3-2\ln 8+\ln 16$?

Answer
VerifiedVerified
551.1k+ views
Hint: We will use the properties of logarithm to solve this question. We will use the quotient rule and product rule of the logarithm which are given as
Product rule – $\ln \left( m\times n \right)=\ln m+\ln n$
Quotient rule – $\ln \left( \dfrac{m}{n} \right)=\ln m-\ln n$

Complete step by step answer:
We have been given an expression $\ln 3-2\ln 8+\ln 16$
We have to simplify the given expression.
We know that logarithm is the inverse function to the exponentiation. We will use the basic properties of logarithm to solve further.
We have $\ln 3-2\ln 8+\ln 16$
Now we know that $\ln {{m}^{n}}=m\ln n$
So we can rewrite the given expression as
$\Rightarrow \ln 3-\ln {{8}^{2}}+\ln 16$
Now, simplifying further we get
$\Rightarrow \ln 3-\ln 64+\ln 16$
Now, we that the product rule of logarithm is given as $\ln \left( m\times n \right)=\ln m+\ln n$
So by applying the product rule in the above expression we get
\[\begin{align}
  & \Rightarrow \left( \ln 3+\ln 16 \right)-\ln 64 \\
 & \Rightarrow \left( \ln 3\times 16 \right)-\ln 64 \\
\end{align}\]
Now, simplifying the obtained equation we get
\[\Rightarrow \ln 48-\ln 64\]
Now, we know that the quotient rule of the logarithm is given as $\ln \left( \dfrac{m}{n} \right)=\ln m-\ln n$
Now, by applying the quotient rule in the above obtained expression we get
\[\Rightarrow \ln \left( \dfrac{48}{64} \right)\]
Now, simplifying the obtained equation we get
\[\Rightarrow \ln \left( \dfrac{3}{4} \right)\]

Hence we get $\ln 3-2\ln 8+\ln 16=\ln \left( \dfrac{3}{4} \right)$.

Note: If the base of logarithm is not given in the question we assume that it is a common logarithm. Logarithm of base is called common logarithm. For example ${{\log }_{10}}36$. Logarithms of base e are called natural logarithms. Avoid calculation mistakes and be careful while applying the properties. If we apply product rule in the terms $\ln 3-\left( \ln 64+\ln 16 \right)$, we will get incorrect answers as the minus sign is there outside the parentheses which change the meaning of complete expression.
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