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

• MHB
• karush
In summary: In-Summary-6.6.61-Kilani-High-School.pdfyou 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

Gold Member
MHB
$\tiny{\textit{6.6.61 Kilani High School}}$
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

x has to be > 10 …

oh yeah I now see that

karush said:
oh yeah I now see that

solve this “modified” equation …

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

skeeter said:
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

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

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

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.

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

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

$\bigg| \dfrac{x-10}{x-6} \bigg|$ 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 $x \ge 10$

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

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

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

DaalChawal said:
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

https://dl.orangedox.com/QS7cBvdKw55RQUbliE

## 1. What is the equation 6.6.61 log(x-10)-log(x-6)=log(2) used for?

The equation 6.6.61 log(x-10)-log(x-6)=log(2) is used to solve for the value of x in a logarithmic function.

## 2. How do I solve the equation 6.6.61 log(x-10)-log(x-6)=log(2)?

To solve this equation, you can use the properties of logarithms to combine the terms on the left side of the equation. Then, you can use the inverse property of logarithms to isolate the variable x and solve for its value.

## 3. Can this equation be solved without a calculator?

Yes, this equation can be solved without a calculator by using the properties of logarithms and simplifying the terms on the left side of the equation. However, a calculator may be helpful in evaluating the final answer.

## 4. What is the domain of this equation?

The domain of this equation is all real numbers greater than 10 and less than 6, since the logarithmic function is undefined for negative and zero values.

## 5. Are there any restrictions on the values of x that can be used in this equation?

Yes, there are restrictions on the values of x that can be used in this equation. The value of x must be greater than 10 and less than 6, as well as positive, since the logarithmic function is undefined for negative and zero values.