If $x = \log 2$ and $y = \log 3$, express $\log 75$ in terms of $x$ and $y$?

Hint: To solve this problem we will have to use the rules of logarithmic function in order to solve this problem. We will use the rules of logarithm to break $\log 75$ into logarithmic functions of $2$ and $3$, that is into, $\log 2$ and $\log 3$. We will then use the required operations to turn $\log 75$ into turns forms of $x$ and $y$. So, let us see how to solve this problem.

Complete step by step answer:
Given that, $x = \log 2$,and $y = \log 3$. We will use, in what follows, the familiar rules of logarithmic function.We have,
$\log 75$
We can write this one like,
$\log 75 = \log \left( {3 \times {5^2}} \right)$
We know, $\log \left( {x \times y} \right) = \log x + \log y$.
So, using this property, we get,
$\log 75 = \log 3 + \log {5^2}$

We know, $\log \left( {{x^y}} \right) = y\log x$,
So, using this property, we get,
$\log 75 = \log 3 + 2\log 5$
Given, $x = \log 2$ and $y = \log 3$, we get,
$y + 2\log 5$
We can write it as,
$y + 2\log \left( {\dfrac{{10}}{2}} \right)$
We know, $\log \left( {\dfrac{x}{y}} \right) = \log x - \log y$.

So, using this property, we get,
$y + 2\left[ {\log 10 - \log 2} \right]$
We know, $\log 10 = 1$,
Therefore,
$ y + 2\left[ {1 - \log 2} \right]$
Now, we are given, $x = \log 2$ and $y = \log 3$, we get,
$ y + 2\left[ {1 - x} \right]$
$\therefore y + 2 - 2x$

Therefore, $\log 75 = y + 2 - 2x$.

Note: The logarithmic functions have values assigned to them for every number greater than $0$. It is known to us that $\log 1 = 0$. For any number $n > 1$, value of $\log n > 0$. And, for every other number $0 < n < 1$, the value of $\log n < 0$. The value of logarithmic functions is not valid for any number less than $0$, that is, for negative real numbers, the value of $\log $ does not exist. We can also say that the domain of logarithmic function is only positive real numbers.