Number of Real Solutions to a logarithmic equation

Question

"How many real solutions does the following equation have?" $$\log_6(3x-26)-\log_6(x+2)=\log_6(-7+x)$$

I tried 2 methods and both gave me $x=6$; $x=2$ so I selected Two Real Solutions, Both Positive however the practice test is telling me the answer is No Real Solutions

Where am I going wrong and is there a quicker method to answering the question than solving the entire equation?

Answer 1

your equation can be simplified to $$\log_6\left(\frac{3x-26}{x+2}\right)=\log_6(-7+x)$$ this can be simplified to $$\frac{3x-26}{x+2}=-7+x$$ can you solve this? it must be $$3x-26>0$$ and $$x+2>0$$ and $$-7+x>0$$ we get $$x=2$$ or $$x=6$$

Answer 2

The answer should be no real solutions. Notice that substituting your answers back in gives log(negative number) on the right hand side of your equation. Negative numbers are not included in the domain of logarithms. Therefore, your answer is no real solution.