MHB 6.6.61 log(x-10)-log(x-6)=log(2)

[karush]

$\tiny{\textit{6.6.61 Kilani High School}}$
Solve for x give exact for\\
$\log{(x-10)}-\log{(x-6)}=\log{2}$
$\begin{array}{rrll}
\textsf{subtraction rule} &\log\left(\dfrac{x-10}{x-6}\right)&=\log{2} \\
\textsf{drop logs} &\dfrac{x-10}{x-6}&=2 \\
&x-10&=2(x-6)=2x-12\\
\textsf{isolate x} &2&=x
\end{array}$

hopefully ok but?
quess we could just put all the logs on r side and set em to zero

 

[skeeter]

x has to be > 10 …

 

[karush]

oh yeah I now see that

 

[skeeter]

oh yeah I now see that

solve this “modified” equation …

$\log|x-10| - \log|x-6| = \log(2)$

 

[karush]

solve this “modified” equation …

$\log|x-10| - \log|x-6| = \log(2)$

$\log|x-10| - \log|x-6| = \log(2)$
$\log\left|\dfrac{x-10}{x-6}\right|= \log(2)$
$\left|\dfrac{x-10}{x-6}\right|= 2$th
$\left|x-10\right|= 2\left|{x-6}\right|$
ok got this
$x=2\quad or \quad x=\dfrac{22}{3}$
or is it an interval

 

[DaalChawal]

Hint: see what happens for $x \gt 10 , 6 \lt x \lt 10$ and $x \lt 6$ and open modulus accordingly.

 

[karush]

what is modulus?
i tried this on a desmom plot but ?

 

[DaalChawal]

I meant mod... you must be knowing that mod breaks in plus and minus or like if $y=|x|$ then for all values of x, y will be positive all the time. This question does not need desmos, try on your own.

 

[karush]

so like if x=11 the +, -, -

 

[DaalChawal]

No No... I'm saying if $|x|=5$ then $x= +5 , -5$ . I guess you have not studied modulus function or its properties yet.

 

[skeeter]

$\bigg| \dfrac{x-10}{x-6} \bigg|$ has two critical values … x = 10 and x = 6

$x \ge 10 \implies \dfrac{x-10}{x-6} \ge 0 \implies \dfrac{x-10}{x-6} = 2 \implies x =2$, however, $x=2$ is not in the interval $x \ge 10$

$6 < x < 10 \implies \dfrac{x-10}{x-6} < 0 \implies -\left(\dfrac{x-10}{x-6}\right) = 2 \implies x = \dfrac{22}{3}$

$x < 6 \implies \dfrac{x-10}{x-6} > 0 \implies \dfrac{x-10}{x-6} = 2 \implies x=2$

 

[DaalChawal]

The interval will be $x \gt 10 $ not $x \ge 10$ because $log 0$ is not defined.

 

[skeeter]

The interval will be $x \gt 10 $ not $x \ge 10$ because $log 0$ is not defined.

you are correct … I was focused on solving $\bigg| \dfrac{x-10}{x-6} \bigg| =2$ and forgot it was the argument of a log function

 

[karush]

https://dl.orangedox.com/QS7cBvdKw55RQUbliE