I can't find in my notes how to do this exactly. please help with how to set this up thank you sooo much

f(x)= log7(2- x+5 0 over x-6)

sorry, i don't know how to type up the fraction

2. Hello
do you mean $f(x)=\log_{7}\left (2-\frac{x+5}{x-6} \right )$.

3. yes, that it was i mean

4. I haven't done this question before in my life, but here goes,

I don't have the tool to do log base thingy, so I'll merely show you the method, and you simply follow the procedures carefully.

Let log base 10 be lg

Showing simple concepts,

lg x* = a
=> x* = 10^a --------- am I right? ----- then what is the domain of lg x*?

as you know the basic log graph, where the range of lg x* = a: (-∞,∞), basically all real values.

Now, back to " x* = 10^a ", where R of lg x* = a: ( -∞, +∞ )

when a tends to -∞, x* = 10^(-∞) = 1/(10^∞) ≈ 0
when a tends to + ∞, x* = 10^(∞) = ∞

Hence, logically, the domain of lg x*: ( 0, ∞ )

Ok, back to business ( Qns ); I will edit the qns a little because I can't make the log 7 thingy.

Qns: lg [2 -(x+5)/(x-6)] = lg A = lg x*

Let [2 -(x+5)/(x-6)] be A

As I have mentioned, the domain of A = domain of x*: ( 0, ∞ )

Meaning, since, " 0 < x* < ∞ ", then " 0 < A < ∞ " = " 0 < [2 -(x+5)/(x-6)] < ∞ "

but here is the problem, we have to make x the subject to make an inequality, right?

Method:

Since (x+5)/(x-6) = 1 + 11/(x-6),

then [2 -(x+5)/(x-6)] = [2 - 1 - 11/(x-6)] = [ 1 - 11/(x-6) ]

Solving the inequality for the LHS first,

0 < 1 - 11/(x-6)
=> -1 < - 11/(x-6)
=> 1 > 11/(x-6)
=> (x-6) > 11
=> x > 17

Now, solving it for the RHS,

1 - 11/(x-6) < ∞
=> -11/(x-6) < ∞ - 1 = ∞
=> (x-6) < 11/∞ ≈ 0
=> x < 6

Hence the domain of x: [ x<6 & x>17 ]

This is answer you seek, I guess; the domain is the same regardless of any log base.

5. Originally Posted by thisisnew
yes, that it was i mean
the domain is,
$D=\left \{ x\in \mathbb{R}\mid x\neq 6,\frac{x-7}{x-6}> 0 \right \}$
therefore,
$D=(-\infty,6 )\cup (17,\infty )$

6. Originally Posted by werepurple
I haven't done this question before in my life, but here goes,

I don't have the tool to do log base thingy, so I'll merely show you the method, and you simply follow the procedures carefully.

Let log base 10 be lg

Showing simple concepts,

lg x* = a
=> x* = 10^a --------- am I right? ----- then what is the domain of lg x*?

as you know the basic log graph, where the range of lg x* = a: (-∞,∞), basically all real values.

Now, back to " x* = 10^a ", where R of lg x* = a: ( -∞, +∞ )

when a tends to -∞, x* = 10^(-∞) = 1/(10^∞) ≈ 0
when a tends to + ∞, x* = 10^(∞) = ∞

Hence, logically, the domain of lg x*: ( 0, ∞ )

Ok, back to business ( Qns ); I will edit the qns a little because I can't make the log 7 thingy.

Qns: lg [2 -(x+5)/(x-6)] = lg A = lg x*

Let [2 -(x+5)/(x-6)] be A

As I have mentioned, the domain of A = domain of x*: ( 0, ∞ )

Meaning, since, " 0 < x* < ∞ ", then " 0 < A < ∞ " = " 0 < [2 -(x+5)/(x-6)] < ∞ "

but here is the problem, we have to make x the subject to make an inequality, right?

Method:

Since (x+5)/(x-6) = 1 + 11/(x-6),

then [2 -(x+5)/(x-6)] = [2 - 1 - 11/(x-6)] = [ 1 - 11/(x-6) ]

Solving the inequality for the LHS first,

0 < 1 - 11/(x-6)
=> -1 < - 11/(x-6)
=> 1 > 11/(x-6)
=> (x-6) > 11
=> x > 17

Now, solving it for the RHS,

1 - 11/(x-6) < ∞
=> -11/(x-6) < ∞ - 1 = ∞
=> (x-6) < 11/∞ ≈ 0
=> x < 6

Hence the domain of x: [ x<6 & x>17 ]

This is answer you seek, I guess; the domain is the same regardless of any log base.
your domain notation doesn't make any sense.