Question

# Solve the following expression:${{\log }_{10}}\left( {{x}^{2}}-x-6 \right)-x={{\log }_{10}}\left( x+2 \right)-4$ .

Hint: First of all separate all the terms containing log. Then use the formula $\log m-\log n=\log \dfrac{m}{n}$. Also use if ${{\log }_{m}}n=a$, then $n={{\left( m \right)}^{a}}$ to find the desired value of x. Take special care of the domain of the logarithmic function.

Here, we have to solve the equation, ${{\log }_{10}}\left( {{x}^{2}}-x-6 \right)-x={{\log }_{10}}\left( x+2 \right)-4$.

Let us consider the equation given in the question,
${{\log }_{10}}\left( {{x}^{2}}-x-6 \right)-x={{\log }_{10}}\left( x+2 \right)-4....\left( i \right)$

By taking the terms containing log to one side and remaining terms to the other side, we get,
${{\log }_{10}}\left( {{x}^{2}}-x-6 \right)-{{\log }_{10}}\left( x+2 \right)=x-4$

We know that $\log m-\log n=\log \dfrac{m}{n}$. By using this in the above equation, we get,
${{\log }_{10}}\left( \dfrac{{{x}^{2}}-x-6}{\left( x+2 \right)} \right)=\left( x-4 \right)....\left( ii \right)$

Now, taking the expression, $E={{x}^{2}}-x-6$

We can also write it as,
$E={{x}^{2}}-\left( 3x-2x \right)-6$
Or, $E={{x}^{2}}-3x+2x-6$

We can write the above expression as,
$E=x\left( x-3 \right)+2\left( x-3 \right)$

By taking $\left( x-3 \right)$ common, we get,
$E=\left( x+2 \right)\left( x-3 \right)$
So, we get, ${{x}^{2}}-x-6=\left( x+2 \right)\left( x-3 \right)$. By substituting the value of ${{x}^{2}}-x-6$ in equation (ii), we get,
${{\log }_{10}}\left[ \dfrac{\left( x+2 \right)\left( x-3 \right)}{\left( x+2 \right)} \right]=\left( x-4 \right)$
By canceling the like terms, we get,
${{\log }_{10}}\left( x-3 \right)=\left( x-4 \right)$
We know that if ${{\log }_{m}}n=a$, then $n={{\left( m \right)}^{a}}$

By using this, we get,
$\left( x-3 \right)={{10}^{\left( x-4 \right)}}....\left( iii \right)$
We know that for ${{\log }_{n}}m$, m should always be greater than zero.
Therefore, for $\log \left( x+2 \right)$, we get,
$\left( x+2 \right)>0$
Or, $x>-2....\left( iv \right)$

Also for, $\log \left( {{x}^{2}}-x-6 \right)$, we get,
$\left( {{x}^{2}}-x-6 \right)>0$
Or, $\left( x+2 \right)\left( x-3 \right)>0$

For x > 3, (x + 2) > 0 and (x â€“ 3) > 0
This means that, (x + 2) (x â€“ 3) > 0
For example, let us take x = 4, then,
(4 + 2) (4 â€“ 3) = 6 > 0
Therefore, $x\in \left( 3,\infty \right)$
For â€“ 2 < x < 3, (x + 2) > 0 and (x â€“ 3) < 0
This means that, (x + 2) (x â€“ 3) < 0
For example, let us take x = 0 then (0 + 2) (0 â€“ 3) = -6 < 0
Therefore, x does not belong to this category.
For x < - 2, (x + 2) < 0 and (x â€“ 3) < 0
This means that, (x + 2) (x â€“ 3) < 0
For example, let us take x = -3, then
(â€“ 3 + 2) (â€“ 3 â€“ 3) = 6 > 0
Therefore, $x\in \left( -\infty ,-2 \right)$
Hence, we get $x\in \left( -\infty ,-2 \right)\cup \left( 3,\infty \right)....\left( v \right)$
Now we know that (x + 2) and $\left( {{x}^{2}}-x-6 \right)$ should both be positive. Hence, we will take the intersection of values of x, we get,
$x\in \left( -\infty ,-2 \right)\cap \left[ \left( -\infty ,-2 \right)\cup \left( 3,\infty \right) \right]$

The only region which is common to both is $\left( 3,\infty \right)$.
So, we get, $x\in \left( 3,\infty \right)$.
Now, to consider equation (iii), that is,
$\left( x-3 \right)={{10}^{\left( x-4 \right)}}$
Now, here we will substitute the value of x > 3
Let us substitute x = 4. So, we get,
$\left( 4-3 \right)={{10}^{\left( 4-4 \right)}}$
$\Rightarrow 1={{10}^{0}}$
$\Rightarrow 1=1$
LHS = RHS
Therefore, x = 4 is the solution of the equation.
Now, in the equation $\left( x-3 \right)={{10}^{\left( x-4 \right)}}$, if we will substitute the value of x > 4, then RHS of the above equation that is ${{10}^{\left( x-4 \right)}}$ will rise very fast (exponentially) like ${{10}^{5}},{{10}^{6}}$ , etc. But the LHS of the above equation will rise very slowly like 2, 3, 4, etc. So, they wonâ€™t be equal for any value of x.
So, we get only one solution and that is x = 4.

Note: In this question, many students just solve the given equation and give the wrong answer. Special care should be taken for the domain of a logarithmic function that if we have ${{\log }_{n}}m$, then â€˜mâ€™ should be greater than zero and â€˜nâ€™ should also be greater than zero and $n\ne 1$. Also, students can cross-check their answer by substituting x = 4 back in the equation given in the question.