Question

How do you solve $\log \left( {x + 2} \right) + \log \left( {x - 1} \right) = 4$?

Answer 1

When one term is raised to the power of another term, the function is called an exponential function, for example $a = {x^y}$ . The inverse of the exponential functions are called logarithm functions, the inverse of the function given in the example is$y = {\log _x}a$ that is a logarithm function. These functions also obey certain rules called laws of the logarithm, using these laws we can write the function in a variety of ways. Using the laws of the logarithm, we can solve the given problem.

Complete step-by-step solution:
Given,
$\log \left( {x + 2} \right) + \log \left( {x - 1} \right) = 4$
We know logarithm product law, that is $\log (x.y) = \log (x) + \log (y)$. applying this we have,
$\Rightarrow \log \left( {\left( {x + 2} \right)\left( {x - 1} \right)} \right) = 4$
$\Rightarrow \log \left( {x(x - 1) + 2(x - 1)} \right) = 4$
$\Rightarrow \log \left( {{x^2} - x + 2x - 2} \right) = 4$
$\Rightarrow \log \left( {{x^2} + x - 2} \right) = 4$
taking antilog we will have,
$\Rightarrow {x^2} + x - 2 = {10^4}$
(because we have logarithm base 10)
$\Rightarrow {x^2} + x - 2 = 10000$
$\Rightarrow {x^2} + x - 2 - 10000 = 0$
$\Rightarrow {x^2} + x - 10002 = 0$
We have a quadratic equation, to solve this we use quadratic formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
On comparing with $a{x^2} + bx + c = 0$ we have $a = 1,b = 1,c = - 10002$, substituting we have,
$\Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(1)( - 10002)} }}{{2(1)}}$
$\Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 40008} }}{2}$
$\Rightarrow x = \dfrac{{ - 1 \pm \sqrt {40009} }}{2}$

Hence the solution of $\log \left( {x + 2} \right) + \log \left( {x - 1} \right) = 4$ is $x = \dfrac{{ - 1 + \sqrt {40009} }}{2}$ and $x = \dfrac{{ - 1 - \sqrt {40009} }}{2}$

Note: To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is $\log (x.y) = \log (x) + \log (y)$. Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. that is $\log \left( {\dfrac{x}{y}} \right) = \log x - \log y$. Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is $\log {x^a} = a\log x$. These are the basic rules we use while solving a problem that involves logarithm function.