-1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Are you attempting to solve the equation (solve for $x$?)? $\endgroup$ – Namaste Nov 29 '16 at 16:03
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $\ln(1) = 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.