Can you help me solve this inequality?

Hi S.A.M. Can you show us where you got stuck? Thanks!

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SucAtMath said:
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If your name is accurate, you probably can't handle this. But then, you probably wouldn't have been assigned it.

Try expressing everything in terms of u = 3^x and v = 5^x. You'll end up with a homogeneous polynomial in two variables. Then see what you can do with it. It's a nice challenge. (This is just the easiest approach I found to describe in a brief hint, not necessarily the best.)
 
Dr.Peterson said:
If your name is accurate, you probably can't handle this. But then, you probably wouldn't have been assigned it.

Try expressing everything in terms of u = 3^x and v = 5^x. You'll end up with a homogeneous polynomial in two variables. Then see what you can do with it. It's a nice challenge. (This is just the easiest approach I found to describe in a brief hint, not necessarily the best.)
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I've got a better plan for it, can I post my solution?
 
canvas said:
I've got a better plan for it, can I post my solution?
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I wouldn't. S.A.M. has to learn to solve it without being shown the answer. If you can give just a hint, go ahead.

If you really need to show your whole method to someone (I found several), you might message one of us.
 
Dr.Peterson said:
I wouldn't. S.A.M. has to learn to solve it without being shown the answer. If you can give just a hint, go ahead.

If you really need to show your whole method to someone (I found several), you might message one of us.
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I sent you a private message, check it out! :p
 
Dr.Peterson said:
I wouldn't. S.A.M. has to learn to solve it without being shown the answer. If you can give just a hint, go ahead.

If you really need to show your whole method to someone (I found several), you might message one of us.
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Or at least wait 24 hours - if S.A.M. does not show any ore work.
 
Hi S.A.M. Can you show us where you got stuck? Thanks!
Otis said:
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I got stuck after simplifying the solution into 3^x and 5^x
I got 3^(3x)+(5^(2x))(3^x)<2*5^(3x) and I assigned 3^x as a and 5^x as b
I got stuck at a^3+ab^2-2b^3<0
 
[math]3^{3x}+5^{3x}=2*5^{3x}[/math]
Dividing every term for [imath]3^{3x}[/imath] yields to

[math]1+\left (\frac{5}{3}\right )^{3x}<2*\left (\frac{5}{3}\right )^{3x} \Rightarrow \left (\frac{5}{3}\right )^{3x} > 1 \Rightarrow x > 0[/math]
 
Eugenio said:
[math]3^{3x}+5^{3x}=2*5^{3x}[/math]
Dividing every term for [imath]3^{3x}[/imath] yields to

[math]1+\left (\frac{5}{3}\right )^{3x}<2*\left (\frac{5}{3}\right )^{3x} \Rightarrow \left (\frac{5}{3}\right )^{3x} > 1 \Rightarrow x > 0[/math]
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Don't give the full solution! Have you not read the thread!? It has not been
24 hours, just for an instance. If you can come back by editing time, at least
hide what you typed.

Edit: It was pointed out to me by a member that I missed the incorrect math made by Eugenio in post # 12 that I referred to.
 
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SucAtMath said:
Hi S.A.M. Can you show us where you got stuck? Thanks!

I got stuck after simplifying the solution into 3^x and 5^x
I got 3^(3x)+(5^(2x))(3^x)<2*5^(3x) and I assigned 3^x as a and 5^x as b
I got stuck at a^3+ab^2-2b^3<0
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Good work so far. That's the homogeneous polynomial I mentioned (every term has degree 3).

A useful trick is to divide by either a^3 or b^3, which turns it into an inequality in one variable, a/b or b/a.
 
lookagain said:
Don't give the full solution! Have you not read the thread!? It has not been
24 hours, just for an instance. If you can come back by editing time, at least
hide what you typed.
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It's been 6 hours. Enough torture I guess. The guy did an effort and the solution to this problem can give him ideas for other problems he is stuck in. Don't be that hard.
 
Good work so far. That's the homogeneous polynomial I mentioned (every term has degree 3).
Dr.Peterson said:
A useful trick is to divide by either a^3 or b^3, which turns it into an inequality in one variable, a/b or b/a.
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I tried but I‘m very hard stuck on this part.
 
Eugenio said:
It's been 6 hours. Enough torture I guess. The guy did an effort and the solution to this problem can give him ideas for other problems he is stuck in. Don't be that hard.
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No, you guessed wrong. The OP was in the process of getting hints, not a solution,
from a helper. He was not done with his effort. It was ongoing. And, you posted
wrong math, which is not helpful. You may want to work on a site that gives
solutions. If you want to give a solution after a while, give it for a similar problem,
and use correct Mathematics.

I reported your post # 12. At the time of the reporting, I did so fast that I did not
get to notice your wrong mathematics to add that reason in the report.

Your response here to me is out of line. It should have been along the line of
"lookagain, I'm newer here about the procedures. I will do better."

The moderators can tell you more.
 
SucAtMath said:
I tried but I‘m very hard stuck on this part.
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Can you show us what you obtained after following the advice of post#14 please?

EDIT: It's worth repeating that the method in post#12 is flawed, so please don't use that, 5^3 = 125 but the original question requires 75
 
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lookagain said:
No, you guessed wrong. The OP was in the process of getting hints, not a solution,
from a helper. He was not done with his effort. It was ongoing. And, you posted
wrong math, which is not helpful. You may want to work on a site that gives
solutions. If you want to give a solution after a while, give it for a similar problem,
and use correct Mathematics.

I reported your post # 12. At the time of the reporting, I did so fast that I did not
get to notice your wrong mathematics to add that reason in the report.

Your response here to me is out of line. It should have been along the line of
"lookagain, I'm newer here about the procedures. I will do better."

The moderators can tell you more.
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I will take the idea of a similar problem. But, why do you say that my math is incorrect? Is there any reason or is it just incorrect?
Hey, if you (or anybody) give me good reasons we can have a nice discussion but if you have aggressive behavior I will ignore you and your posts.
 
Cubist said:
Can you show us what you obtained after following the advice of post#14 please?

EDIT: It's worth repeating that the method in post#12 is flawed, so please don't use that, 5^3 = 125 but the original question requires 75
Click to expand...

Ok, that´s the reason. But that's not incorrect math. It's a mistake. Thanks for pointing it.
The idea is to convert the original inequality into an exponential inequality with the same base.
Please, correct me if I'm wrong. The original equation can be written as

[math](3^3)^x+(3*5^2)^x<10*5^{3x}*5^{-1}[/math][math]3^{3x}+3^x*5^{2x}<2*5^{3x}[/math]
Now we can apply the idea of Dr.Peterson and turn it into an inequality in one variable, a/b or b/a.
 
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