Show that log 360 = 2log3 + 3log2 + log5.

Hint: For solving this logarithmic related problem, we will generally use some of the properties of logarithmic function such as ${{\log }_{a}}M\times N={{\log }_{a}}M+{{\log }_{a}}N$ and ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$ . Then we will arrange the terms of logarithmic in such a manner so that we can replace the complex logarithmic expression to the simplest form using standard results.

Complete step by step answer:
Now, in question we are given the expression 2log3 + 3log2 + log5 and we are asked to show that on solving we get log360.
Now, we know that property of log which is ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$,
Similarly we can say that, in 2log3 is equal to $2\log 3=\log {{3}^{2}}$ and for, 3log2, we will have $3\log 2=\log {{2}^{3}}$ and let log5 be remain as it is.
So, adding all above values we have,
 $\log {{3}^{2}}+\log {{2}^{3}}+\log 5$,
Also, we know that ${{\log }_{a}}M\times N={{\log }_{a}}M+{{\log }_{a}}N$
So, we will have expression$\log \left( {{3}^{2}}\times {{2}^{3}}\times 5 \right)$, for $\log {{3}^{2}}+\log {{2}^{3}}+\log 5$
On simplifying, we get
$\log \left( 9\times 8\times 5 \right)$, as ${{3}^{2}}=9,{{2}^{3}}=8$
On solving we get, log 360
Hence, $log\text{ }360\text{ }=\text{ }2log3\text{ }+\text{ }3log2\text{ }+\text{ }log5$

Note: Firstly, we must know all the properties of logarithmic functions such as ${{\log }_{N}}M=\dfrac{{{\log }_{a}}M}{{{\log }_{a}}N}$ , ${{\log }_{a}}M\times N={{\log }_{a}}M+{{\log }_{a}}N$ , ${{\log }_{a}}M=\dfrac{1}{{{\log }_{M}}a}$, ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$, ${{\log }_{a}}{{a}^{a}}=a$. The logarithmic of two numbers to the same base leads same value if and only if both numbers are same .One of the most basic trick to solve logarithmic functions is to re – arrange the order of base value and variable value such that you left with some expression to which you can replace with some standard results.