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C3 rearranging equation

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
Reply 1
Your question is not clear enough
Reply 2
how?
Reply 3
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.
Reply 4
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.
Reply 5
Original post by steve2005
You might like to add some brackets.


you're right, should be: e^-0.8x-(1/3-2x)=0
Reply 6
So its like: e0.8x13+2x=0?e^{-0.8x}-\frac13 + 2x = 0?
(edited 11 years ago)
Reply 7
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)..
Reply 8
is this clearer?.. e^-0.8x-(1/(3-2x))=0
Reply 9
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
Reply 10
sorry, its definitely e^0.8x-(1/(3-2x))=0

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

new to this..


Unparseable latex formula:

\displaystyle e^-^{0.8x}- \frac{1}{3-2x} = 0

Finally:rolleyes:

And you are after something in the form

x=pln(32x)\displaystyle x = p \ln (3-2x)

Just for you:smile:

Is it

Unparseable latex formula:

\displaystyle e^-^0^.^8^x



or

Unparseable latex formula:

\displaystyle e^0^.^8^x



you are after:confused:

Log laws

aln(x)=ln(xa)\displaystyle a \ln (x) = \ln (x^a)
(edited 11 years ago)
Original post by Shao Kahn
sorry, its definitely e^0.8x-(1/(3-2x))=0

new to this..





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Is this it ?
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
Unparseable latex formula:

\displaystyle e^0^.^8^x- \frac{1}{3-2x} = 0



(P.S. SubAtomic - to do e0.8xe^{0.8x} 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: 132x=(32x)1\dfrac{1}{3-2x} = (3-2x)^{-1}
Reply 14
Original post by steve2005



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Is this it ?


yes! but with e^-0.8x!
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)
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.
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




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(edited 11 years ago)
Original post by steve2005
I think p = 1.25




Uploaded with ImageShack.us


Yep that's what I got - but she said the answer says -1.25 which I was saying is wrong :smile:
Reply 19
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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