0
$\begingroup$

$$ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}.$$

I made the graph using wolfram alpha it is giving answer as 6. But how to solve it algebraically? base of logarithm is 2.

Tried using taking Lcm but then two different log terms are getting formed.

$\endgroup$
1
$\begingroup$

$$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2} = \frac{5}{12}$$

Rewrite the logs using $$\log_a b^c = c\log_a b$$

and factor. Then, simplify both sides.

$$\frac{1}{2}\cdot\frac{1}{\log_2\vert x-2\vert}+\frac{1}{2}\cdot\frac{1}{\log_2\vert x+2\vert} = \frac{5}{12}$$

$$\frac{1}{2}\bigg(\frac{1}{\log_2\vert x-2\vert}+\frac{1}{\log_2\vert x+2\vert}\bigg) = \frac{5}{12}$$

$$\frac{1}{\log_2\vert x-2\vert}+\frac{1}{\log_2\vert x+2\vert} = \frac{5}{6}$$

$$\frac{\log_2\vert x-2\vert+\log_2\vert x+2\vert}{\log_2\vert x-2\vert\cdot\log_2\vert x+2\vert} = \frac{5}{6}$$

Set $\color{blue}{a = \log_2\vert x-2\vert}$ and $\color{purple}{b = \log_2\vert x+2\vert}$.

$$\frac{\color{blue}{a}+\color{purple}{b}}{\color{blue}{a}\color{purple}{b}} = \frac{5}{6}$$

Can you take it on from here? (Hint: Solve for possible values of $a$ and $b$. Then, plug in $\color{blue}{\log_2\vert x-2\vert = a}$ and $\color{purple}{\log_2\vert x+2\vert = b}$ and check for any extraneous solutions, in case there are any.)

$\endgroup$
  • $\begingroup$ no information given on a, b ? so how can we solve for a, b ? hit and trial?? $\endgroup$ – maveric Nov 13 '18 at 15:32
  • $\begingroup$ Trial and error is the quickest way. You can conclude both $a+b > 0$ and $ab > 0$ or both $a+b < 0$ and $ab < 0$. For the first case, both $a > 0$ and $b > 0$, which gets $a = 3$ and $b = 2$ and vice-versa. For the second case, either $a > 0$ and $b < 0$ or $a < 0$ and $b > 0$, getting $a = 1$ and $b = -6$ and vice-versa. $\endgroup$ – KM101 Nov 13 '18 at 15:40
  • $\begingroup$ Anyway, there could possibly be a simpler way. This is what I thought of immediately though. $\endgroup$ – KM101 Nov 13 '18 at 15:46
  • 1
    $\begingroup$ Careful! $ \log_{2}(x-2)^2 = 2\log_{2}|x-2| $ $\endgroup$ – medicu Nov 13 '18 at 17:43

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.