Explanation by means of an example.
log(x^{2} - 2) = log (-x)The general method to solve an equation consists in repeatedly replacing the equation by an equivalent equation.
To solve the equation, we are tempted to omit the log-function. But the following equivalence is false!
log(x^{2} - 2) = log (-x) <=> x^{2} - 2 = -xNotice that 1 is a solution of x^{2} - 2 = -x but it is not a solution of log(x^{2} - 2) = log (-x). So, the equations log(x^{2} - 2) = log (-x) and x^{2} - 2 = -x are not equivalent.
It is clear that following expression is correct.
log(x^{2} - 2) = log (-x) => x^{2} - 2 = -xIf two numbers have the same logarithm, then those numbers are equal. All solutions of the equation on the left are solutions of the equation on the right but not vice versa.
But we will solve logarithmic equations by omitting the log-function on both sides. But every time we must be aware that, by omitting the log-function, the new equation can have more solutions than the previous one. We really do not want these extra solutions. We call them parasitic solutions.
There are several ways to eliminate the parasitic solutions. The easiest way is testing the solutions to the given equation. The solutions that do not satisfy the original equation are the parasitic solutions. They are deleted.
Let's use this procedure to the equation log (x^{2} - 2) = log (-x).
log(x^{2} - 2) = log (-x) => x^{2} - 2 = -xThis gives a quadratic equation with roots 1 and -2. We test these two values to the original equation and we see that only -2 is a solution. The value 1 is a parasitic solution and must be deleted.
First, we bring the equation in the form log A = log B and we omit the log-function.
log(x^{2}+1) = log(10) => x^{2}+1 = 10We find 3 and -3 as the solutions of the last equation. We test these two values to the original equation and we see that they both satisfy. Here, there are no parasitic solutions.
First, we bring the equation in the form log A = log B and we omit the log-function.
log(x-2) = log(10) + log(x+2) log(x-2) = log 10.(x+2) x-2 = 10.(x+2)We find -22/9 as the solution of the last equation. We test this value to the original equation and we see that it does not satisfy. -22/9 is a parasitic solution. The given equation has no solution.
First, we bring the equation in the form log A = log B and we omit the log-function.
log(x-3)(x+1) = log(2x+2) (x-3)(x+1) = (2x+2)We find -1 and 5 as the solutions of the last equation. 5 is a solution. -1 is a parasitic solution and must be deleted.
First, we bring the equation in the form log A = log B and we omit the log-function.
log(x-1)^{2} = log (2x^{2}+x-3) (x-1)^{2} = 2x^{2}+x-3We find 1 and -4 as the solutions of the last equation. Both values are parasitic. The given equation has no solution.
First, we bring the equation in the form log A = log B and we omit the log-function.
log (log(x^{2}+1) + 9) = log(10) log(x^{2}+1) + 9 = 10 log(x^{2}+1) = 1 log(x^{2}+1) = log(10) x^{2}+1 = 10We find 3 and -3 as the solutions of the last equation. We test these two values to the original equation and we see that they both satisfy.
First, we bring the equation in the form log A = log B and we omit the log-function.
log(x+2) + log(x^{2}) = log(x^{2}-2x) log( (x+2).x^{2} ) = log(x^{2}-2x) (x+2).x^{2} = x^{2}-2xThe only real solution is 0. The value is parasitic. The given equation has no solution.