Question

How does one solve $\log (x - 3) + \log x = 1$?

Answer 1

For simplifying the original equation , firstly used logarithm property $\log a + \log b = \log (ab)$ then take base ten 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.

Complete solution step by step:
We have
$\log (x - 3) + \log x = 1 \\\ or \\\ \log ((x - 3)x) = 1 \\\$
By assuming the base of the logarithm to be $10$ ,then take the base $10$ exponential of both sides of the equation,
${10^{{{\log }_{10}}((x - 3)x)}} = {10^1}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get ,
$(x - 3)x = 10$
Simplify the equation,
$\Rightarrow {x^2} - 3x - 10 = 0$
For finding roots of the original equation, we have to use quadratic formula i.e.,
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now identify $a,b,c$ from the original equation given below,
$\Rightarrow {x^2} - 3x - 10 = 0$
$\Rightarrow a = 1 \\\ \Rightarrow b = - 3 \\\ \Rightarrow c = - 10 \\\$
Put these values into the formula of finding the roots of quadratic equations,
$x = \dfrac{{3 \pm \sqrt {{3^2} - 4*1*( - 10)} }}{{2*1}}$
After simplifying and by evaluating exponents and square root of the above equation we get the following simplified expression,
$x = \dfrac{{3 \pm 7}}{2}$
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{{3 + 7}}{2} \\\ and \\\ \Rightarrow {x_2} = \dfrac{{3 - 7}}{2} \\\$
Simplify and then isolate $x$ to find its corresponding solutions!
$\Rightarrow {x_1} = 5 \\\ and \\\ \Rightarrow {x_2} = - 2 \\\$
Now recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero.
Therefore, in our original equation $\log (x - 3) + \log x = 1$,
Here,
$(x - 3) > 0 \\\ and \\\ (x) > 0 \\\$
Now after evaluating both the values, $- 2$ is rejected because it is less than zero. While $5$ is greater than zero and $5 - 3 > 0$ .
Therefore, we have our solution i.e., $5$ .
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.,
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

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