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# How do you solve $\log x + \log (x - 3) = 1$ ?

Last updated date: 21st Feb 2024
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Hint: In this particular question we need to use basic logarithmic properties to simplify the equation. Then we need to further solve the equation and get the desired answer.

Complete step by step solution:
In the above question, it is given that,
$\log x + \log (x - 3) = 1$
(Since $\log a + \log b = \log (ab)$ )
Using the above stated property we get,
$\Rightarrow \log (x(x - 3)) = 1$
Taking antilog on both sides of the equation we get,
$\Rightarrow anti\log (\log (x(x - 3))) = anti\log 1$
$\Rightarrow x(x - 3) = {10^1}$
On solving the above equation we get a quadratic equation
$\Rightarrow {x^2} - 3x = 10$
Subtracting 10 from both sides of the equation
$\Rightarrow {x^2} - 3x - 10 = 0$
Now solve the above quadratic equation for x
$\Rightarrow {x^2} - 5x + 2x - 10 = 0$
$\Rightarrow x(x - 5) + 2(x - 5) = 0$
$\Rightarrow (x + 2)(x - 5) = 0$
This implies that either $x = - 2$or $x = 5$

As log can not be a negative value therefore the $x = - 2$ is rejected and hence $x = 5$ is the required solution to the above logarithmic equation.

Note:
Remember to recall the basic logarithmic properties to solve the above question. Note that
$\log x = 1 \\ \Rightarrow x = {10^1} \\ \Rightarrow x = 10 \\$
The basic logarithmic algebra includes the following properties:
$\log a + \log b = \log (ab) \\ \log a - \log b = \log \left( {\dfrac{a}{b}} \right) \\ \log {a^b} = b\log a \\ {\log _a}a = 1 \\$