How does one solve $\log (x - 15) + \log x = 2$ ?

Hint:For simplifying the original equation , firstly used logarithm property $\log a + \log b = \log (ab)$ then take base $10$ exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the equation and lastly used quadratic formula for finding the roots.

Formula used:
We used logarithm properties i.e.,
$\log a + \log b = \log (ab)$ ,
${b^{{{\log }_b}a}} = a$ and
We also used quadratic formula i.e.,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

Complete step by step answer:
We have
\[\Rightarrow log(x - 15) + log(x) = 2 \\
\Rightarrow log(x(x - 15)) = 2\]
By assuming the base of the logarithm to be $10$ ,then take the base $10$ exponential of both sides of the equation,
\[ \Rightarrow {10^{log(x(x - 15))}} = {10^2}\]
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we get
\[ \Rightarrow (x(x - 15)) = 100\]
Simplify the equation,
\[ \Rightarrow {x^2} - 15x - 100 = 0\]

For finding roots of the original equation, we have to use quadratic formula i.e.,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now identify $a,b,c$from the original equation given below,
\[{x^2} - 15x - 100 = 0 \\
\Rightarrow a = 1 \\
\Rightarrow b = - 15 \\
\Rightarrow c = - 100 \]
Put these values into the formula of finding the roots of quadratic equations,
$x = \dfrac{{ - ( - 15) \pm \sqrt {{{( - 15)}^2} - 4\times 1\times ( - 100)} }}{{2 \times 1}}$
After simplifying and by evaluating exponents and square root of the above equation we get the following simplified expression,
$ \Rightarrow x = \dfrac{{15 \pm 25}}{2}$

To find the roots of the equations , separate the particular equation into its corresponding parts : one part with the plus sign and the other with the minus sign.
$\Rightarrow {x_1} = \dfrac{{15 + 25}}{2} \\
\Rightarrow {x_2} = \dfrac{{15 - 25}}{2} \\ $
Simplify and then isolate $x$ to find its corresponding solutions!
$\Rightarrow {x_1} = 20, \\
\Rightarrow {x_2} = - 5$
Now recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero. Therefore, in our original equation $\log (x - 15) + \log x = 2$ ,
$\Rightarrow (x - 15) > 0, \\
\Rightarrow x > 0 \\ $
Now after evaluating both the values, $ - 5$ is rejected because it is less than zero. While $20$ is greater than zero and $20 - 15 > 0$.

Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. For finding the roots of the equation we use a quadratic formula. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one .