Solving log inequalities

[ghi]

Solve the inequality.

log(x-2) + log(9-x) < 1

I know the answer is 2<x<4 or 7<x<9 but I don't know how to get there.

 

[pka]

Solve the inequality.
log(x-2) + log(9-x) < 1
I know the answer is 2<x<4 or 7<x<9 but I don't know how to get there.

Note because of domain issues the final answer must be \(\displaystyle 2<x<9\).

Also \(\displaystyle \log(a)<1\text{ if and only if }0<a<10.\)

So with the domain restrictions solve
\(\displaystyle (x-2)(9-x)<10.\)

 

[lookagain]

Note because of domain issues the final answer must be
\(\displaystyle > > > ... \ \ 2<x<9 \ \ ... < < < ?\).


Also \(\displaystyle \log(a)<1\text{ if and only if }0<a<10.\)


So with the domain restrictions solve
\(\displaystyle (x-2)(9-x)<10.\)

pka,

in your last inequality,
these values of x, for instance, do not work:


\(\displaystyle x = 4, \ \ \ \ \ \ (2)(5) = 10\)

\(\displaystyle x = 4.5, \ \ \ (2.5)(4.5) > 10\)

\(\displaystyle x = 5, \ \ \ \ \ \ (3)(4) > 10\)

\(\displaystyle x = 5.5, \ \ \ (3.5)(3.5) > 10\)

\(\displaystyle x = 6, \ \ \ \ \ \ (4)(3) > 10\)

\(\displaystyle x = 6.5, \ \ \ (4.5)(2.5) > 10\)

\(\displaystyle x = 7, \ \ \ \ \ \ (5)(2) = 10\)


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pka,

did you mean for the word \(\displaystyle "in"\) to be
between "must be" and "2 < x < 9?"

 

[pka]

pka,
did you mean for the word \(\displaystyle "in"\) to be
between "must be" and "2 < x < 9?"

No, I meant what I said.
The domain issue means \(\displaystyle x\in (2,9)\).
The inequality means \(\displaystyle (x-2)(9-x)<10\).
Taken together their intersection gives the reported answer.

You should know that the word "in" for an open interval means between its endpoints.

 

[lookagain]

\(\displaystyle ** \)Note because of domain issues the final answer must be \(\displaystyle 2<x<9\).

Also \(\displaystyle \log(a)<1\text{ if and only if }0<a<10.\)

So with the domain restrictions solve
\(\displaystyle (x-2)(9-x)<10.\)

** Then you didn't type what you meant. This sentence means that you are claiming
the "final answer" is 2 < x < 9, which it isn't. That is your goof.

Now, you would have been correct if you had typed something along the lines of:

Note: Because of domain issues, the final answer must be contained in (or must be a subset of)
2 < x < 9.


The phrase "the final answer must be 2 < x < 9" does not mean "the final answer must
in/within/part of/etc. of 2 < x < 9."


I have no problem calling you out on that. Just reread what you originally typed.
The actual meaning in your statement is inconsistent with what you intended.

No, I meant what I said. ----> No, you did not say what you meant.
The domain issue means \(\displaystyle x\in (2,9)\).
The inequality means \(\displaystyle (x-2)(9-x)<10\).
Taken together their intersection gives the reported answer.

*** You should know that the word "in" for an open interval means between its endpoints.

No, you left out the word "in" where it belongs. The omission of "in" immediately to the left of[i/]
the intervachanged the meaning.


What you had reads as:

"...because of domain issues the final answer must be the interval
x is greater than 2 and is less than 9."