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\[{{\log }_{10}}\left( {{x}^{2}}-x-6 \right)-x={{\log }_{10}}\left( x+2 \right)-4\] .

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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

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