
How do you solve $ \ln 12 - \ln (x - 1) = \ln (x - 2) $ ?
Answer
544.8k+ views
Hint: The inverse of the exponential functions are called logarithm functions and is of the form $ y = {\log _x}a $ that is a logarithm function. When the base of the function is equal to e, then it is called a natural logarithm function. Certain rules called laws of the logarithm are obeyed by these functions, and using these laws we can write the function in a variety of ways. The function given in the question is a natural logarithm function, as the base of both these functions is the same, we can apply the logarithm laws in the given question.
Complete step-by-step answer:
There are three laws of the logarithm, one of addition, one of subtraction and the other to convert logarithm functions to exponential functions. In the given question, the two logarithm functions are in subtraction with each other so we have used the law for subtraction. Then we will simplify the expression obtained and find the value of x by using the appropriate method.
We are given that $ \ln 12 - \ln (x - 1) = \ln (x - 2) $
We know that
$
\log x - \log y = \log \dfrac{x}{y} \\
\Rightarrow \ln 12 - \ln (x - 1) = \ln \dfrac{{12}}{{x - 1}} \;
$
Using the above result in the given equation, we get –
$ \ln \dfrac{{12}}{{x - 1}} = \ln (x - 2) $
We also know that –
$
if,\,{\log _n}x = {\log _n}y \\
\Rightarrow x = y \;
$
So,
$
\dfrac{{12}}{{x - 1}} = x - 2 \\
\Rightarrow 12 = (x - 2)(x - 1) \\
\Rightarrow {x^2} - x - 2x + 2 = 12 \\
\Rightarrow {x^2} - 3x - 10 = 0 \;
$
On the factorization of the obtained equation, we get –
$
{x^2} - 5x + 2x - 10 = 0 \\
\Rightarrow x(x - 5) + 2(x - 5) = 0 \\
\Rightarrow (x + 2)(x - 5) = 0 \\
\Rightarrow x + 2 = 0,\,x - 5 = 0 \\
\Rightarrow x = - 2,\,x = 5 \;
$
Hence, if $ \ln 12 - \ln (x - 1) = \ln (x - 2) $ then $ x = - 2 $ or $ x = 5 $ .
So, the correct answer is “ $ x = 5 $ ”.
Note: While applying the laws of the logarithm, we must note that the base of the logarithm functions involved should be the same in all the calculations. The equation obtained after removing the log is a polynomial equation; it has a degree 2 as the highest power of x is 2, so it is a quadratic equation and can be solved by factorization as shown.
Complete step-by-step answer:
There are three laws of the logarithm, one of addition, one of subtraction and the other to convert logarithm functions to exponential functions. In the given question, the two logarithm functions are in subtraction with each other so we have used the law for subtraction. Then we will simplify the expression obtained and find the value of x by using the appropriate method.
We are given that $ \ln 12 - \ln (x - 1) = \ln (x - 2) $
We know that
$
\log x - \log y = \log \dfrac{x}{y} \\
\Rightarrow \ln 12 - \ln (x - 1) = \ln \dfrac{{12}}{{x - 1}} \;
$
Using the above result in the given equation, we get –
$ \ln \dfrac{{12}}{{x - 1}} = \ln (x - 2) $
We also know that –
$
if,\,{\log _n}x = {\log _n}y \\
\Rightarrow x = y \;
$
So,
$
\dfrac{{12}}{{x - 1}} = x - 2 \\
\Rightarrow 12 = (x - 2)(x - 1) \\
\Rightarrow {x^2} - x - 2x + 2 = 12 \\
\Rightarrow {x^2} - 3x - 10 = 0 \;
$
On the factorization of the obtained equation, we get –
$
{x^2} - 5x + 2x - 10 = 0 \\
\Rightarrow x(x - 5) + 2(x - 5) = 0 \\
\Rightarrow (x + 2)(x - 5) = 0 \\
\Rightarrow x + 2 = 0,\,x - 5 = 0 \\
\Rightarrow x = - 2,\,x = 5 \;
$
Hence, if $ \ln 12 - \ln (x - 1) = \ln (x - 2) $ then $ x = - 2 $ or $ x = 5 $ .
So, the correct answer is “ $ x = 5 $ ”.
Note: While applying the laws of the logarithm, we must note that the base of the logarithm functions involved should be the same in all the calculations. The equation obtained after removing the log is a polynomial equation; it has a degree 2 as the highest power of x is 2, so it is a quadratic equation and can be solved by factorization as shown.
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