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I found the answers, but the back of the book has something different?

  • #1

Homework Statement



The problem is

ln ( 2 - 5x) > 2

Homework Equations



The answers for the back of the book is x < (2-e^2) / 5

The Attempt at a Solution



Here's my attempt

ln ( 2 - 5x ) > 2

= ln 2 - ln 5x > 2

= - ln 5x > 2 - ln 2

= x < (2- ln 2) / -ln 5

Is there a way I can simplify this even more guys. thanks for all your help so far
 

Answers and Replies

  • #2
danago
Gold Member
1,122
4


Here's my attempt

ln ( 2 - 5x ) > 2

= ln 2 - ln 5x > 2


= - ln 5x > 2 - ln 2

= x < (2- ln 2) / -ln 5

Is there a way I can simplify this even more guys. thanks for all your help so far


Have a look at that line ive bolded. Perhaps have a look at your logarithm rules again, and see if you can see what you have done wrong. The answer the book has is infact correct.
 
  • #3
Have a look at that line ive bolded. Perhaps have a look at your logarithm rules again, and see if you can see what you have done wrong. The answer the book has is infact correct.
Thank you so much for over looking my work.

Let's see, I've think i've simplied it correctly this time, but is there a way to get it really bone dried simplied? Thanks for your help

I've got

= e^2 > 2- 5x

= e^2 - 2 > -5 x

= ( e^2 - 2) / -5 < x

I can't get the signs to match for some reason and negative and positive stuff. I'm so close my good sir
 
  • #4
danago
Gold Member
1,122
4
Your answer is correct now. It is exactly the same as the books answer, but they have just multiplied both the top and bottom of the fraction by -1, just to make it look a bit neater.

It is basically the same idea as -5/-2 = 5/2
 
  • #5
Your answer is correct now. It is exactly the same as the books answer, but they have just multiplied both the top and bottom of the fraction by -1, just to make it look a bit neater.
You're correct sir, but I still can't get my greater than or less than sign to match up. i'm mulitplying top and bottom by -1 so that's twice making my < sign stay the same. Thanks for over looking my work my kind sir, but just missing one part of the puzzle : )
 
  • #6
danago
Gold Member
1,122
4
You're correct sir, but I still can't get my greater than or less than sign to match up. i'm mulitplying top and bottom by -1 so that's twice making my < sign stay the same. Thanks for over looking my work my kind sir, but just missing one part of the puzzle : )
Oh sorry didnt even realise you had the inequality sign the wrong way around. How did you you go from ln (2-5x) > 2 to e^2 > 2- 5x? It seems you have got the inequality sign the wrong way round here.
 
  • #7
Oh sorry didnt even realise you had the inequality sign the wrong way around. How did you you go from ln (2-5x) > 2 to e^2 > 2- 5x? It seems you have got the inequality sign the wrong way round here.
I dont understand, i wrote down the problem correctly, and i just change the original problem to better see it to:

log (base e) (2-5x ) > 2

Then I change it to the exponential form

e^2 > 2-5x

Did I miss something with the change of signs?
 
  • #8
danago
Gold Member
1,122
4
If ln (2-5x) > 2, then that should imply that 2-5x > e^2, not the other way around.
 
  • #9
If ln (2-5x) > 2, then that should imply that 2-5x > e^2, not the other way around.
But I didn't write down the problem wrong. I really don't understand where I missed the sign change.
 
  • #10
danago
Gold Member
1,122
4
You went from ln (2-5x) > 2 to e^2 > 2-5x. All im saying is that you got the inequality sign the wrong way around in that first step.
 
  • #11
You went from ln (2-5x) > 2 to e^2 > 2-5x. All im saying is that you got the inequality sign the wrong way around in that first step.
I see I have to keep an eye on that, if I'm having the same confuzino here. BTW, thank you for your time, can't say that enough, im stressing about a big math final.

Like for example on antoher one im stuck

e ^( x^2 - 4x) > = e ^ 5

I take the natural log of both sides and it becomes

X^2 - 4x >= 5

Then I subtract 5

x^2 - 4x - 5

and I factor

(x - 5 ) ( x + 1 ) >= 0

so x => 5 or x >= -1

But the back of the book says x <= -1 or x >=5

These signs are killing me, my friend
 
  • #12
danago
Gold Member
1,122
4
All good up until (x - 5 ) ( x + 1 ) >= 0.

What we are essentially saying there is that the product of two numbers (x-5) and (x+1) is positive. That means they must be the same sign; either both positive, or both negative.

Lets look at the case where they are both positive:
x-5>0 Therefore x>5
x+1>0 Therefore x>-1

We need when they are BOTH positive, so that only occurs when x>5.

If you do the same and find when they are BOTH negative, you will see that x<-1.

Does that make sense to you?
 
  • #13
All good up until (x - 5 ) ( x + 1 ) >= 0.

What we are essentially saying there is that the product of two numbers (x-5) and (x+1) is positive. That means they must be the same sign; either both positive, or both negative.

Lets look at the case where they are both positive:
x-5>0 Therefore x>5
x+1>0 Therefore x>-1

We need when they are BOTH positive, so that only occurs when x>5.

If you do the same and find when they are BOTH negative, you will see that x<-1.

Does that make sense to you?
danago, Thank you so much! yes that does make sense to me, because a positive times a positive equals a positive and a negative times a negative is a positive. I see why that would be a solutkion

Is there anyway I can repay you for all your help?
 
  • #14
danago
Gold Member
1,122
4
danago, Thank you so much! yes that does make sense to me, because a positive times a positive equals a positive and a negative times a negative is a positive. I see why that would be a solutkion

Is there anyway I can repay you for all your help?
Exactly right :)

Just send the cheque to... haha nah no need to repay me in any way, i like helping out when i can. Good luck with your final!
 
  • #15
Can't thank you enough, but i'll be sure to help other members in return with what i can. Thanks again
 

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