Prove the given logarithmic expression $\log \dfrac{9}{14}+\log \dfrac{35}{24}-\log \dfrac{15}{16}=0$ using appropriate logarithm formula.

Hint: We should know the logarithm formula for product of a,b is $\log (ab)=\log (a)+\log (b)$ and similarly also we should know that the division of a,b is $\log \left( \dfrac{a}{b} \right)=\log (a)-\log (b)$.According to these formulae the given question has to be proved to zero.

Complete step by step answer:
We have given in the problem that ,
$\log \dfrac{9}{14}+\log \dfrac{35}{24}-\log \dfrac{15}{16}=0$⟶equation(1)
Let us consider the L.H.S part of the given problem
 ⟹ $\log \dfrac{9}{14}+\log \dfrac{35}{24}-\log \dfrac{15}{16}$
Since,$\log \left( \dfrac{a}{b} \right)=\log (a)-\log (b)$ according to this we get
⟹$[\log (9)-\log (14)]+[\log (35)-\log (24)]-[\log (15)-\log (16)]$
We can expand and write this as follows
⟹$[\log (3\times 3)-\log (7\times 2)]+[\log (7\times 5)-\log (6\times 4)]-[\log (5\times 3)-\log (4\times 4)]$
Similarly we have the product of a,b in logarithm formula is $\log (ab)=\log (a)+\log (b)$
Now on solving according the above formula we get as follows
⟹ $\begin{align}
  & \log (3)+\log (3)-\log (7)-\log (2)+\log (7)+\log (5)-\log (6) \\
 & -\log (4)-\log (5)-\log (3)+\log (4)+\log (4) \\
\end{align}$
{Since((−)×(−)=+)&((+)×(+)=+)&((−)×(+)=−) properties of operators}
The opposite sign terms with equal values will vanishes,the resultant terms are shown below
⟹\[\log 3-\log 2-\log 6+\log 4\]
Which can be written as,
⟹$\log 3-\log 2-\log 3-\log 2+\log 2+\log 2$
On simplifying we get the result as
⟹0
And now the opposite sign term are vanished totally which is equal to zero that R.H.S part of the equation(1)
∴$\log \dfrac{9}{14}+\log \dfrac{35}{24}-\log \dfrac{15}{16}=0$
Hence proved..

Note:
The student may go wrong in simplifying process.The adding and multiplying combinations of
 positive and negative terms can cause confusion and so care must be taken .We must remember that Two ‘pluses’ make a plus,two ‘minuses’ make a plus .A plus and a minus make a minus.