Question & Answer
QUESTION

Evaluate; $2\log 3-\dfrac{1}{2}\log 16+\log 12$.

ANSWER Verified Verified
Hint: Here, first convert $2\log 3$ into $\log {{3}^{2}}$ and$2\log 3-\dfrac{1}{2}\log 16+\log 12$ into $\log 4$ by the identity $\log {{a}^{b}}=b\log a$ and then write all the values in $2\log 3-\dfrac{1}{2}\log 16+\log 12$. After the substitution we also have to apply the identities:
$\begin{align}
  & \log a+\log b=\log ab \\
 & \log a-\log b=\log \dfrac{a}{b} \\
\end{align}$

Complete step-by-step answer:
Here, we have to find the value of $2\log 3-\dfrac{1}{2}\log 16+\log 12$.
We have, $2\log 3-\dfrac{1}{2}\log 16+\log 12$ …. (1)
Now, first consider $2\log 3-\dfrac{1}{2}\log 16+\log 12$.
We know by the logarithmic identity that:
$\log {{a}^{b}}=b\log a$
By applying this identity we will get:
$2\log 3=\log {{3}^{2}}$
Now, by taking the square of 3 which is 9 we will get:
$2\log 3=\log 9$ …… (2)
Now, consider the term $2\log 3-\dfrac{1}{2}\log 16+\log 12$.
Here, also we have to apply the identity:
$\log {{a}^{b}}=b\log a$
Hence, by applying this we will get:
$\dfrac{1}{2}\log 16=\log {{16}^{\dfrac{1}{2}}}$
We know that ${{16}^{\dfrac{1}{2}}}$ is the square root of 16 which is 4, thus we obtain:
 $\dfrac{1}{2}\log 16=\log \sqrt{16}$
$\dfrac{1}{2}\log 16=\log 4$ ……(3)
Now, by substituting equation (2) and equation (3) in equation (1) we obtain:
$2\log 3-\dfrac{1}{2}\log 16+\log 12=\log 9-\log 4+\log 12$
Now, by rearranging the terms we will get:
$2\log 3-\dfrac{1}{2}\log 16+\log 12=\log 9+\log 12-\log 4$
By the logarithmic identity we can write:
$\log a+\log b=\log ab$
By applying the above identity we get:
$\begin{align}
  & 2\log 3-\dfrac{1}{2}\log 16+\log 12=\log (9\times 12)-\log 4 \\
 & 2\log 3-\dfrac{1}{2}\log 16+\log 12=\log 108-\log 4 \\
\end{align}$
Now, we also have another identity,
$\log a-\log b=\log \dfrac{a}{b}$
Next, by applying this identity we get:
$2\log 3-\dfrac{1}{2}\log 16+\log 12=\log \dfrac{108}{4}$
Now, by cancelling 108 by 4 we get 27. Hence our equation becomes:
$2\log 3-\dfrac{1}{2}\log 16+\log 12=\log 27$

Note: Here, you can also do this by converting $\log 12$ into $\log (4\times 3)$. After that, apply the identity $\log a+\log b=\log ab$, you will get it as $\log 4+\log 3$. In the next step, cancel $\log 4$ with $-\log 4$. At last you will get it as $2\log 3+\log 3=3\log 3$. After that by applying the identity you will get$\log {{3}^{3}}=\log 27$.