Question

How do you solve ${\log _5}(x - 1) + {\log _5}(x - 2) - {\log _5}(x + 6) = 0$?

Answer 1

Hint:Use log properties to solve the given equation
The very first step to solve these types of problems is to use the addition and subtraction log properties. We know that $\log a + \log b = \log (ab)$ and $\log a - \log b = \log \dfrac{a}{b}$. So, after using the given properties on the above question, we will get a much simplified version of the question. To move forward we will use the $\log a - \log b = \log \dfrac{a}{b}$ property and then move forward to equate the argument part inside the equation to 0. Solving the equation we will get our final answer.

Complete step by step solution:
The given question we have is ${\log _5}(x - 1) + {\log _5}(x - 2) - {\log _5}(x + 6) = 0$
Now, we will use a simple property of log, and that is
$\log a - \log b = \log \dfrac{a}{b}$
So, applying this rule to the question, the modified version of the question will be:-
$
{\log _5}(x - 1) + \left[ {{{\log }_5}(x - 2) - {{\log }_5}(x + 6)} \right] = 0 \\
{\log _5}(x - 1) + \left[ {{{\log }_5}\left( {\dfrac{{x - 2}}{{x + 6}}} \right)} \right] = 0 \\
$
Now, we will use another log property which states that:-
$\log a + \log b = \log (ab)$
So, when we use this property, our modified version of the question will be:-
$
{\log _5}(x - 1) + \left[ {{{\log }_5}\left( {\dfrac{{x - 2}}{{x + 6}}} \right)} \right] = 0 \\
\Rightarrow {\log _5}\left[ {\left( {x - 1} \right) \times \left( {\dfrac{{x - 2}}{{x + 6}}} \right)} \right] = 0 \\
\Rightarrow {\log _5}\left[ {\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{x + 6}}} \right] = 0
\\
\Rightarrow {\log _5}\left[ {\dfrac{{{x^2} - 2x - x + 2}}{{x + 6}}} \right] = 0 \\
\Rightarrow {\log _5}\left[ {\dfrac{{{x^2} - 3x + 2}}{{x + 6}}} \right] = 0 \\
$
To further simplify this step, we will use another property which states that:-
If,
$
{\log _b}a = x \\
\Rightarrow {b^x} = a \\
$
So, if we use this property on the above question, we will get:-
\[
{\log _5}\left[ {\dfrac{{{x^2} - 3x + 2}}{{x + 6}}} \right] = 0 \\
\Rightarrow {5^0} = \dfrac{{{x^2} - 3x + 2}}{{x + 6}} \\
\Rightarrow 1 = \dfrac{{{x^2} - 3x + 2}}{{x + 6}} \\
\Rightarrow x + 6 = {x^2} - 3x + 2 \\
\]
Subtracting x from both the sides we will get:-
\[
x + 6 = {x^2} - 3x + 2 \\
x + 6 - x = {x^2} - 3x + 2 - x \\
6 = {x^2} - 4x + 2 \\
\]
Subtracting 2 from both the sides we will get:-
\[
6 = {x^2} - 4x + 2 \\
6 - 2 = {x^2} - 4x + 2 - 2 \\
4 = {x^2} - 4x \\
\]
Subtracting 4 from both the sides we will get:-
\[
4 = {x^2} - 4x \\
4 - 4 = {x^2} - 4x - 4 \\
{x^2} - 4x - 4 = 0 \\
\]
To solve this step we will use the quadratic formula method \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here,
a=1
b=-4 and c=-4
so,
\[
x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)( - 4)} }}{{2(1)}} \\
x = \dfrac{{4 \pm \sqrt {16 + 16} }}{2} \\
x = \dfrac{{4 \pm \sqrt {32} }}{2} \\
x = \dfrac{{4 \pm \sqrt {16 \times 2} }}{2} \\
x = \dfrac{{4 \pm 4\sqrt 2 }}{2} \\
x = \dfrac{{2\left( {2 \pm 2\sqrt 2 } \right)}}{2} \\
x = 2 \pm 2\sqrt 2 \\
\]

Note: You must not simply add and subtract the arguments inside the log to solve these problems. Logs don’t work that way. For addition and subtraction of logs, you must use the above mentioned formulas. Logs are special terms in mathematics and hence, their addition, subtraction is quite different from others. If you do it the normal way, you will end up getting the wrong answers.