Question

How do you condense ${{\log }_{4}}\left( 20 \right)-{{\log }_{4}}\left( 45 \right)+{{\log }_{4}}\left( 144 \right)$?

Answer 1

Hint: The given expression in the above question consists of the logarithmic terms. So we can use the logarithmic properties in order to condense all of the three logarithmic terms into a single logarithmic term. For this, we can use the property \[\log A-\log B=\log \left( \dfrac{A}{B} \right)\] on the first two terms of the expression ${{\log }_{4}}\left( 20 \right)-{{\log }_{4}}\left( 45 \right)+{{\log }_{4}}\left( 144 \right)$. Then we can use the logarithm property $\log A+\log B=\log \left( AB \right)$ in order to condense it completely. Then, we can simplify the condensed logarithm term obtained according to the base of the logarithm given. For this, we have to use the logarithm property ${{\log }_{a}}\left( {{a}^{m}} \right)=m$ to get a simplified form of the given expression.

Complete step by step solution:
Let us consider the expression given in the above question as
$ E={{\log }_{4}}\left( 20 \right)-{{\log }_{4}}\left( 45 \right)+{{\log }_{4}}\left( 144 \right)$
We know from the properties of the logarithm function that \[\log A-\log B=\log \left( \dfrac{A}{B} \right)\]. Therefore, we can condense the first two logarithm terms of the above expression as
$\Rightarrow E={{\log }_{4}}\left( \dfrac{20}{45} \right)+{{\log }_{4}}\left( 144 \right)$
Simplifying the fractional argument, we get
$\Rightarrow E={{\log }_{4}}\left( \dfrac{4}{9} \right)+{{\log }_{4}}\left( 144 \right)$
Now, applying the logarithm property $\log A+\log B=\log \left( AB \right)$ on the above expression, we get
\[\begin{align}
  & \Rightarrow E={{\log }_{4}}\left( \dfrac{4}{9}\times 144 \right) \\
 & \Rightarrow E={{\log }_{4}}\left( 4\times 16 \right) \\
 & \Rightarrow E={{\log }_{4}}64 \\
\end{align}\]
Now, we know that $64={{4}^{3}}$. Substituting this above, we get
$\Rightarrow E={{\log }_{4}}\left( {{4}^{3}} \right)$
From the properties of logarithm, we also know that ${{\log }_{a}}\left( {{a}^{m}} \right)=m$. So the above expression finally gets simplified to
$\Rightarrow E=3$
Hence the given expression is condensed and is simplified as $3$.

Note: We can also separate the highest power of the base $4$ from each of the logarithm terms. For this, we can write the given expression as ${{\log }_{4}}\left( 4\times 5 \right)-{{\log }_{4}}\left( 45 \right)+{{\log }_{4}}\left( {{4}^{2}}\times 9 \right)$. Applying the logarithmic property $\log \left( AB \right)=\log A+\log B$ on the first and the third terms, we can condense the given expression.