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# C3 rearranging equation

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1. Hi,
Stuck on ex4c q6c:

Show that the equation e^0.8x-1/3-2x=0 can be written in the form
x=p ln(3-2x), stating the value of p.

Can anyone help? thanks
2. Your question is not clear enough
3. how?
4. you need to get the exponential term on the left and the other stuff on the right.
then think how to change into a non-exponential expression on the left.
5. (Original post by Shao Kahn)
Hi,
Stuck on ex4c q6c:

Show that the equation e^0.8x-1/3-2x=0 can be written in the form
x=p ln(3-2x), stating the value of p.

Can anyone help? thanks
You might like to add some brackets.
6. (Original post by steve2005)
You might like to add some brackets.
you're right, should be: e^-0.8x-(1/3-2x)=0
7. So its like:
8. (Original post by the bear)
you need to get the exponential term on the left and the other stuff on the right.
then think how to change into a non-exponential expression on the left.
right, i think i'm close

e^-0.8x-(1/3-2x)=0

e^-0.8x=(1/3-2x)

ln(-0.8x)=(1/3-2x)

ln(3-2x)=1/-0.8x

ln(3-2x)=-1.25x

i know i'm close because in the answers p=-1.25,
i just can't get it into the form x=p ln(3-2x)..
9. is this clearer?.. e^-0.8x-(1/(3-2x))=0
10. (Original post by Shao Kahn)
right, i think i'm close

e^-0.8x-(1/3-2x)=0

e^-0.8x=(1/3-2x)

ln(-0.8x)=(1/3-2x)

ln(3-2x)=1/-0.8x

ln(3-2x)=-1.25x

i know i'm close because in the answers p=-1.25,
i just can't get it into the form x=p ln(3-2x)..
taking logs is the correct way to proceed but

ln(e-0.8x) is just -0.8x... you must also take the log of the RHS
11. sorry, its definitely e^0.8x-(1/(3-2x))=0

new to this..
12. (Original post by Shao Kahn)
sorry, its definitely e^0.8x-(1/(3-2x))=0

new to this..
Finally

And you are after something in the form

Just for you

Is it

or

you are after

Log laws

13. (Original post by Shao Kahn)
sorry, its definitely e^0.8x-(1/(3-2x))=0

new to this..

Is this it ?
14. (Original post by Shao Kahn)
sorry, its definitely e^0.8x-(1/(3-2x))=0

new to this..
Right okay so it's

(Original post by SubAtomic)
(P.S. SubAtomic - to do you can just do e^{0.8x} with no need to do e^0^.^8^x)

So, Shao Kahn. As before, take the fraction to the right hand side. Natural log both sides.

Hint:
15. (Original post by steve2005)

Is this it ?
yes! but with e^-0.8x!
16. (Original post by Shao Kahn)
yes! but with e^-0.8x!
In that case, the answer is wrong (if the answer is in fact p = -1.25)
17. (Original post by Shao Kahn)
right, i think i'm close

e^-0.8x-(1/3-2x)=0

e^-0.8x=(1/3-2x)

ln(-0.8x)=(1/3-2x)

ln(3-2x)=1/-0.8x

ln(3-2x)=-1.25x

i know i'm close because in the answers p=-1.25,
i just can't get it into the form x=p ln(3-2x)..
Here is where you went wrong.

You should have gotten -0.8x = ln(1/3-2x) instead.
18. (Original post by hassi94)
In that case, the answer is wrong (if the answer is in fact p = -1.25)
I think p = 1.25

19. (Original post by steve2005)
I think p = 1.25

Yep that's what I got - but she said the answer says -1.25 which I was saying is wrong
20. I know this post was awhile ago but:

f(x)=e0.8x - 1/(3-2x)

e0.8x - 1/(3-2x) = 0

e0.8x = 1/(3-2x)

e-0.8x = (3-2x)/1

-0.8x = ln(3-2x)

x= 1/(-0.8)ln(3-2x)

therefore x=-1.25ln(3-2x)

so P= -1.25

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