How do you solve $\log (x - 1) + \log (x + 1) = 2\log (x + 2)$ ?
Answer
604.2k+ views
Hint: In this question, we are given two logarithmic functions equal to each other. So to solve this question, we must know the definition of the logarithm function and the laws of the logarithm. Logarithm functions are the inverse of the exponential functions, they are of the form ${\log _a}b$ where “a” is the base of the function, when we are not given the base of the function, so we take it as 10. As the base of all the logarithm functions involved is the same so by using the laws and properties of the logarithm, we will first simplify the left-hand side and the right-hand side and then we will remove the logarithm function. The equation that is obtained now will be a polynomial equation that can be solved using the appropriate method.
Complete step by step answer:
We know that, $\log a + \log b = \log ab$ and $a\log b = \log {b^a}$ , we get
$ \Rightarrow \log (x - 1)(x + 1) = \log {(x + 2)^2}$
Now, we know that $(a - b)(a + b) = {a^2} - {b^2}$ and ${(a + b)^2} = {a^2} + {b^2} + 2ab$
So,
$ \Rightarrow \log ({x^2} - 1) = \log ({x^2} + 4 + 4x)$
When $\log a = \log b$ , we get $a = b$
$
\Rightarrow {x^2} - 1 = {x^2} + 4 + 4x \\
\Rightarrow {x^2} - {x^2} - 4x = 4 + 1 \\
\Rightarrow - 4x = 5 \\
\Rightarrow x = - \dfrac{5}{4} \\
$
When we plug in the obtained value in the given equation, we get
$ \Rightarrow \log ( - \dfrac{9}{4}) + \log ( - \dfrac{1}{4}) = 2\log (\dfrac{3}{8})$
But for $\log a$ to exist $a > 0$
Hence there are no solutions for the equation $\log (x - 1) + \log (x + 1) = 2\log (x + 2)$.
Note: When the unknown term in an algebraic expression is raised to some non-negative integer power, the expression is said to be a polynomial equation. The equations obtained in the parenthesis of the log are polynomial equations of degree two; but when we solve them, the equation becomes a polynomial equation of degree 1. We know that the number of solutions of a polynomial equation is equal to its degree, so the equation has only one solution that is $ - \dfrac{5}{4}$.
Complete step by step answer:
We know that, $\log a + \log b = \log ab$ and $a\log b = \log {b^a}$ , we get
$ \Rightarrow \log (x - 1)(x + 1) = \log {(x + 2)^2}$
Now, we know that $(a - b)(a + b) = {a^2} - {b^2}$ and ${(a + b)^2} = {a^2} + {b^2} + 2ab$
So,
$ \Rightarrow \log ({x^2} - 1) = \log ({x^2} + 4 + 4x)$
When $\log a = \log b$ , we get $a = b$
$
\Rightarrow {x^2} - 1 = {x^2} + 4 + 4x \\
\Rightarrow {x^2} - {x^2} - 4x = 4 + 1 \\
\Rightarrow - 4x = 5 \\
\Rightarrow x = - \dfrac{5}{4} \\
$
When we plug in the obtained value in the given equation, we get
$ \Rightarrow \log ( - \dfrac{9}{4}) + \log ( - \dfrac{1}{4}) = 2\log (\dfrac{3}{8})$
But for $\log a$ to exist $a > 0$
Hence there are no solutions for the equation $\log (x - 1) + \log (x + 1) = 2\log (x + 2)$.
Note: When the unknown term in an algebraic expression is raised to some non-negative integer power, the expression is said to be a polynomial equation. The equations obtained in the parenthesis of the log are polynomial equations of degree two; but when we solve them, the equation becomes a polynomial equation of degree 1. We know that the number of solutions of a polynomial equation is equal to its degree, so the equation has only one solution that is $ - \dfrac{5}{4}$.
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