Solve (log x)^2 = log x
Also, when will their be a possibility that an equation will have an extraneous solution? Like if you got two solutions for a log equation, will a negative solution for x automatically be extraneous?
if log x = 1
then the statement is true
so if we are using base 10 logs
then x = 10
(logx)^2 - logx = 0
logx(logx-1) = 0
logx=0 at x=1
logx=1 at x=e assuming natural logs
negative will indeed be extraneous, since log(x) is not defined for x <= 0.
However, if you have something like log(2x-7) then you have to discard any solution where 2x-7 <= 0, not just x<=0.
To solve the equation (log x)^2 = log x, we can follow these steps:
Step 1: Let y = log x.
The equation becomes: y^2 = y.
Step 2: Rearrange the equation:
y^2 - y = 0.
Step 3: Factorize the left side of the equation:
y(y - 1) = 0.
Step 4: Apply the zero-product property:
Set each factor equal to zero:
y = 0 or y - 1 = 0.
Step 5: Solve for y in each equation:
For y = 0: log x = 0.
x = 10^0.
x = 1.
For y - 1 = 0: log x = 1.
x = 10^1.
x = 10.
So, the solutions to the equation (log x)^2 = log x are x = 1 and x = 10.
Regarding your second question, the occurrence of extraneous solutions depends on the nature of the equation and the mathematical operations involved. In logarithmic equations, it is possible to have extraneous solutions, particularly when you apply operations that may introduce non-permissible values.
For example, consider the equation log(x - 3) + log(x + 2) = log x. When solving this equation, you would need to consider that the arguments of logarithms cannot be negative. If, after solving, you obtain a solution that results in a negative argument for any of the logarithms, it would be considered an extraneous solution.
However, it is important to note that not every negative solution is extraneous. It depends on the specific equation and the conditions it presents. Therefore, it is crucial to check the solutions obtained against the original equation to identify any extraneous solutions.