Question

If all possible solutions to $\log_4{(3-x)} + \log_{0.25}{(3+x)} = \log_4{(1-x)} + \log_{0.25}{(2x+1)}$ are found, then there will be

• A. only one prime solution
• B. two real solutions
• C. no real solution
• D. none of these

Answer 1

Answer: (D)

none of these

Answer 2

The domain of the equation is $x \in (-1/2, 1)$. The logarithmic equation simplifies to $\log_4{\left(\frac{3-x}{3+x}\right)} = \log_4{\left(\frac{1-x}{2x+1}\right)}$. Equating arguments gives $\frac{3-x}{3+x} = \frac{1-x}{2x+1}$, which leads to the quadratic equation $x^2 - 7x = 0$. The potential solutions are $x=0$ and $x=7$. Only $x=0$ lies within the domain $(-1/2, 1)$, making it the sole real solution. Since $x=0$ is not a prime number, there are not two real solutions, nor no real solution, option (D) "none of these" is correct.