[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

[ReTurd]

Your question is not clear enough

[Shao Kahn]

how?

[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.

[steve2005]

(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.

[Shao Kahn]

(Original post by steve2005)
You might like to add some brackets.

you're right, should be: e^-0.8x-(1/3-2x)=0

[mr tim]

So its like:

[Shao Kahn]

(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)..

[Shao Kahn]

is this clearer?.. e^-0.8x-(1/(3-2x))=0

[the bear]

(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

[Shao Kahn]

sorry, its definitely e^0.8x-(1/(3-2x))=0

new to this..

[SubAtomic]

(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

[steve2005]

(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 ?

[Intriguing Alias]

(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:

[Shao Kahn]

(Original post by steve2005)



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Is this it ?

yes! but with e^-0.8x!

[Intriguing Alias]

(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)

[justinawe]

(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.

[steve2005]

(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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[Intriguing Alias]

(Original post by steve2005)
I think p = 1.25




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

[MEH:)]

I know this post was awhile ago but:

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

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

e^0.8x = 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