Question

How do you condense $2\log 3+3\log 2-\log 6$?

Answer 1

Hint: The given expression is to be condensed and for this we will be using properties of logarithmic function. We will first use \[x\log a=\log {{a}^{x}}\] to get the expression in the uniform form. We will then use the formulae: \[\log a+\log b=\log ab\] and \[\log a-\log b=\log \dfrac{a}{b}\]. Using these formulae, we will further simplify the terms and hence, we will get the condensed form of the given expression.

Complete step by step solution:
According to the given question, we have an expression which we have to condense in the simplest way possible. We will be using the various logarithm properties to carry out the condensation.
The expression we have is,
$2\log 3+3\log 2-\log 6$-----(1)
We will be first using the property,
\[x\log a=\log {{a}^{x}}\]
Using the above formula, we get the expression as,
\[\Rightarrow \log {{3}^{2}}+\log {{2}^{3}}-\log 6\]----(2)
We are now considering the terms \[\log {{3}^{2}}\] and \[\log {{2}^{3}}\],
So, we will be using the formula of the logarithm function, which is,
\[\log a+\log b=\log ab\]
And we get,
\[\Rightarrow \log ({{3}^{2}}{{.2}^{3}})-\log 6\]
We know that, \[{{3}^{2}}=9\] and \[{{2}^{3}}=8\]. Solving the above expression, we get,
\[\Rightarrow \log (9.8)-\log 6\]
\[\Rightarrow \log (72)-\log 6\]-----(3)
The above equation is similar to a formula of logarithm function, which is,
\[\log a-\log b=\log \dfrac{a}{b}\]
Using the above formula, in the equation (3), we get,
\[\Rightarrow \log \dfrac{72}{6}\]
Dividing 72 by 6, we get the expression as,
\[\Rightarrow \log 12\]
Therefore, the condensed form of the given expression is:
\[2\log 3+3\log 2-\log 6=\log 12\]

Note: From equation (2), we chose the terms \[\log {{3}^{2}}\] and \[\log {{2}^{3}}\] to proceed on, but we could also have chosen \[\log {{2}^{3}}\] and \[\log 6\] first, so we have,
Proceeding from equation (2) onwards,
\[\Rightarrow \log {{3}^{2}}+\log {{2}^{3}}-\log 6\]
We know that, \[\log a-\log b=\log \dfrac{a}{b}\],
We will now consider \[\log {{2}^{3}}\] and \[\log 6\], we get,
\[\Rightarrow \log {{3}^{2}}+\log \dfrac{{{2}^{3}}}{6}\]
On solving, we get,
\[\Rightarrow \log 9+\log \dfrac{8}{6}\]
Now, we will use the formula, we get,
\[\log a+\log b=\log ab\]
So, we get,
\[\Rightarrow \log \dfrac{9.8}{6}\]
We get,
\[\Rightarrow \log \dfrac{72}{6}\]
\[\Rightarrow \log 12\]
Therefore, we get the condensed expression, that is,
\[2\log 3+3\log 2-\log 6=\log 12\]