Problem: Log function with common base equation

Question

ln(x-2)-ln(x+9)=ln(x-1)-ln(x+14)

I dropped the ln's and tried solving by using the quotient property but I think im doing the fractions wrong.

can someone please explain how to get the answer? (solving for x)

Thank you

Answer 1

$\ln(x-2)-\ln(x+9)=\ln(x-1)-ln(x+14) \Rightarrow \ln \left(\frac{x-2}{x+9}\right)=\ln \left(\frac{x-1}{x+14}\right)$, but $\ln (x)$ is bijective and so $$\frac{x-2}{x+9}=\frac{x-1}{x+14}$$ and solving that we have: $$(x-2)(x+14)=(x-1)(x+9) \Rightarrow x=19/4$$ and testing that solution at the original equation we see that it holds.

Answer 2

I'm approaching this within the context of the natural log, until we have a clear equation with which to solve for $x$.

$$\ln(x-2)-\ln(x+9) =\ln(x-1)-\ln(x+14) $$

$$\iff \ln\left(\frac{x-2}{x+9}\right) = \ln\left(\frac{x-1}{x+14}\right)$$

$$\iff \ln\left(\frac{x-2}{x+9}\right) - \ln\left(\frac{x-1}{x+14}\right)= 0$$

$$\iff \ln\left(\frac{\frac{x-2}{x + 9}}{\frac{x-1}{x+14}}\right) = 0$$

$$\iff \ln\left(\frac{(x-2)(x+14)}{(x-1)(x+9)}\right)= 0$$

Noting that $\ln(f(x)) = 0 \iff f(x)= 1$, we have $f(x) = \frac{g(x)}{h(x)} = 1$, meaning we must have $$(x-2)(x+14) = (x-1)(x+9)\tag{solve for x}$$

to arrive at $\ln(1) = 0$