Find the domain of 1/(2^x + 3^x - 5^x)

[Math Tutor]

Find the domain of the expression:
[tex]\frac{1}{2^x + 3^x - 5^x}[/tex]

[Guest]

All "x" not equal to one (1).

[Math Tutor]

Can you prove it, please?

[Guest]

Yes. If We have x=1 so we get [tex]\frac{1}{2^{1}+3^{1}-5^{1} } =\frac{1}{ 2+3-5} =\frac{1}{0 }[/tex], but we know that you can not divide by zero, so we take this: All "x" not equal to one (1).

[Math Tutor]

How can we be sure that there is no other solutions?

[Math Tutor]

The answer is [tex]x\ne1[/tex] but it is interesting how you will prove it.

[Guest]

It`s easy.
If we have[tex]x>1[/tex], so we can see that [tex]2^{x}+3^{x} <[/tex]that [tex]5^{x}[/tex]. Yes? For exaple: [tex]x=3[/tex]; [tex]2^{3}+3^{3}=8+27=35[/tex], and [tex]5^{3}=125.[/tex] [tex]125>35[/tex] and while we take [tex]x>>>>1;[/tex] [tex]5^{x}>>>>2^{x}+3^{x}[/tex]
If we have [tex]x<1[/tex], so we can see that [tex]2^{x}+3^{x} >[/tex] that [tex]5^{x}[/tex]. Example the same.
[tex]x=1[/tex] - it`s gold centre and ONLY here we can not divide by zero(result of [tex]x=1[/tex]). It`s all. Sorry for my errors, because i don`t know math`s termins on English)

[Math Tutor]

Thank you for the elegant solution!

[shemet]

It`s easy.
If we have[tex]x>1[/tex], so we can see that [tex]2^{x}+3^{x} <[/tex]that [tex]5^{x}[/tex]. Yes? For exaple: [tex]x=3[/tex]; [tex]2^{3}+3^{3}=8+27=35[/tex], and [tex]5^{3}=125.[/tex] [tex]125>35[/tex] and while we take [tex]x>>>>1;[/tex] [tex]5^{x}>>>>2^{x}+3^{x}[/tex]
If we have [tex]x<1[/tex], so we can see that [tex]2^{x}+3^{x} >[/tex] that [tex]5^{x}[/tex]. Example the same.
[tex]x=1[/tex] - it`s gold centre and ONLY here we can not divide by zero(result of [tex]x=1[/tex]). It`s all. Sorry for my errors, because i don`t know math`s termins on English)

Solution?! Is it a joke?! You must PROVE that [tex]2^{x}+3^{x}<5^{x}[/tex]!!!...

[Guest]

I know, that you VERY clever, but x can be as "+", as "-"

[Guest]

If [tex]x>1, 2^{x}+3^{x}<5^{x} \Leftrightarrow (\frac{2}{5 })^{x}+(\frac{3}{5 })^{x}<1[/tex]. Really [tex]\frac{2}{5 }<1,x>1\Leftrightarrow (\frac{2}{5 })^{x}<(\frac{2}{5 })^{1}[/tex], and [tex]\frac{3}{5 }<1,x>1\Leftrightarrow (\frac{2}{5 })^{x}<(\frac{3}{5 })^{1}[/tex].
Then [tex](\frac{2}{5 })^{x}+(\frac{3}{5 })^{x}<\frac{2}{ 5}+\frac{3}{ 5} = 1[/tex].
Analogous, if [tex]x<1[/tex]then [tex]2^{x}+3^{x}>5^{x}[/tex].
This is a PROOF!!!

[Guest]

How do u proof this ?
It is not so clear.
[tex]\frac{3}{5 }<1,x>1\Leftrightarrow (\frac{2}{5 })^{x}<(\frac{3}{5 })^{1}[/tex]

[shemet]

[tex]\frac{3}{5 }<1,x>1\Leftrightarrow (\frac{3}{5 })^{x}<(\frac{3}{5 })^{1}[/tex]. This is right!