Solving logarithmic eqn with $x^2$

Question

Full question:

Solve the equation: $$\log_5{(x+1)} = \log_{25}{(4x^2)} - \log_2{0.5}^3$$

So,

$\log_5{(x+1)} = \log_{25}{(4x^2)} + \log_2{2}^3$

$\log_5{(x+1)} = \log_{25}{(4x^2)} + 3$

$\log_5{(x+1)} = \frac{1}{2}\log_{5}{(4x^2)} + 3$

$\log_5{(x+1)} = \log_{5}{(2x)} + \log_55^3$

So,

$x+1 = 250x$

$249x = 1$

$x= \frac{1}{249} $

This is the answer provided by the examiner but I have been playing around with a graphing application and it shows that there is a negative answer of $-\frac{1}{251}$. Why is it so?

Answer 1

You lost one solution when you assumed that $4x^2 = (2x)^2$. It could just as well be $(-2x)^2$, leading to $\log_5(x+1) = \log_5(-2x) + 3$. This is, of course, only meaningful when $-1<x<0$, but it does lead to the solution $x = -\frac{1}{251}$

Answer 2

First we have to find a 'domain' for this equation: $$x+1>0 \wedge 4x^2>0$$ Thus the domain is $$D=(-1,\infty)\backslash\{0\}$$

Now from $$\log_5(x+1)=\frac{1}{2}\log_5(4x^2)+3$$ we can have two cases:

1. $\log_5(x+1)=\log_5(2x)+3$ - the domain in this cases is $D_1 = (0,\infty)$ and the result is provided by You ($x=\frac{1}{249}\in D_1$)
2. $\log_5(x+1)=\log_5(-2x)+3$ - in this case $D_2=(-1,0)$ and the result is $x=-\frac{1}{251}\in D_2$

Both results are in $D$, thus the result of yor equation is $$x\in\left\{-\frac{1}{251}, \frac{1}{249}\right\}$$