Question

How do you condense ${{\log }_{5}}\left( 240 \right)-{{\log }_{5}}\left( 75 \right)-{{\log }_{5}}\left( 80 \right)$?

Answer 1

Hint: In the expression given in the above question, we have three logarithmic terms. For condensing it, we need to write a single logarithmic term equivalent to the three logarithmic terms given in the expression ${{\log }_{5}}\left( 240 \right)-{{\log }_{5}}\left( 75 \right)-{{\log }_{5}}\left( 80 \right)$. For this, we need to use the properties of the logarithm functions. By using the logarithm property $\log A-\log B=\log \left( \dfrac{A}{B} \right)$, we can condense the three logarithmic terms given into one term. Then we can simplify the condensed logarithmic term further by using the properties $\log \left( \dfrac{1}{A} \right)=-\log A$ and ${{\log }_{a}}a=1$.

Complete step by step solution:
Let us write the logarithmic expression given in the above question as
$\Rightarrow E={{\log }_{5}}\left( 240 \right)-{{\log }_{5}}\left( 75 \right)-{{\log }_{5}}\left( 80 \right)$
From the properties of the logarithm functions, we know that $\log A-\log B=\log \left( \dfrac{A}{B} \right)$. Applying this property on the first two terms, we can write the above expression as
\[\Rightarrow E={{\log }_{5}}\left( \dfrac{240}{75} \right)-{{\log }_{5}}\left( 80 \right)\]
Simplifying the fraction inside the first logarithmic term, we get
\[\Rightarrow E={{\log }_{5}}\left( \dfrac{16}{5} \right)-{{\log }_{5}}\left( 80 \right)\]
Again applying the logarithmic property $\log A-\log B=\log \left( \dfrac{A}{B} \right)$, we can write the above expression as
\[\Rightarrow E={{\log }_{5}}\left( \dfrac{16}{5\times 80} \right)\]
Simplifying the argument of the logarithmic term, we get
\[\Rightarrow E={{\log }_{5}}\left( \dfrac{1}{25} \right)\]
Now, we know that $\log \left( \dfrac{1}{A} \right)=-\log A$. So we can write
\[\Rightarrow E=-{{\log }_{5}}25\]
We know that $25={{5}^{2}}$. Putting this above we get
\[\Rightarrow E=-{{\log }_{5}}{{5}^{2}}\]
We know that $\log \left( {{a}^{m}} \right)=m\log a$. Applying this on the above expression, we can write
\[\Rightarrow E=-2{{\log }_{5}}5\]
Finally, we know the logarithmic property ${{\log }_{a}}a=1$. So the above expression becomes
\[\begin{align}
  & \Rightarrow E=-2\left( 1 \right) \\
 & \Rightarrow E=-2 \\
\end{align}\]
Hence, the given expression is condensed to $-2$.

Note: We can also simplify each of the three logarithmic terms by factorising their arguments. For example, we can write the first argument as $240=5\times 48$, the second argument as $75={{5}^{2}}\times 3$, and the third argument as $80=5\times 16$. Then using the logarithm property $\log \left( AB \right)=\log A+\log B$ on each of the logarithmic terms, we condense the given expression.