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Reply 1
you are going with this the wrong way, by exact, they mean as a surd, or a fraction, you use ln function in this
Reply 2
desertwarrior
e^(2x) = 10x

how do you solve

homework says, find exact solutions..
i have found one solution using iteration but can't find the other (there are 2..)

how do you find the exact solutions?


The title says e2x=10e^{2x}=10 and your post says e2x=10xe^{2x}=10x Which is correct?
Reply 3
steve2005
The title says e2x=10e^{2x}=10 and your post says e2x=10xe^{2x}=10x Which is correct?

sorry!!

it's

e^(2x) = 10x
Reply 4
I don't think there are any exact solutions.

If you're really interested, look up the Lambert W-function.
Reply 5
DFranklin
I don't think there are any exact solutions.

If you're really interested, look up the Lambert W-function.

I think that's what wolfram alpha used.. thanks

is there any way to use iteration to find the 2nd solution?
Reply 6
desertwarrior
e^(2x) = 10x

how do you solve

homework says, find exact solutions..
i have found one solution using iteration but can't find the other (there are 2..)

how do you find the exact solutions?


Since you are asked for EXACT solutions I think there is an error in the question.
Reply 7
if it = 10 it would be easy, just use natural logs.

But 10x seems much harder than C3?
Reply 8
mfc20
if it = 10 it would be easy, just use natural logs.

But 10x seems much harder than C3?

yea, weird o.0 use ln on both sides....
Ok, you seem to have a misconception that an answer that is accurate to a large number of decimal places is exact. It isn't.

xn+1=e2xn10x_{n+1}=\frac{e^{2x_n}}{10}

xn+1=ln10xn2x_{n+1}=\frac{\ln {10 x_n}}{2}

Try both with x1=1x_1 = 1
Reply 10
Mr M
Ok, you seem to have a misconception that an answer that is accurate to a large number of decimal places is exact. It isn't.

xn+1=e2xn10x_{n+1}=\frac{e^{2x_n}}{10}

xn+1=ln10xn2x_{n+1}=\frac{\ln {10 x_n}}{2}

Try both with x1=1x_1 = 1

Ah thanks Mr M!!! That's what i've been working on, but I get one solution converging to 1.2713.........
From the shape of the y=e^2x and y=10x graphs I would assume that there surely should be 2 solutions.. but iteration can only find one from my understanding..

am I forgetting something,
P.s. repped you thanks
desertwarrior
Ah thanks Mr M!!! That's what i've been working on, but I get one solution converging to 1.2713.........
From the shape of the y=e^2x and y=10x graphs I would assume that there surely should be 2 solutions.. but iteration can only find one from my understanding..

am I forgetting something,
P.s. repped you thanks


You said the question required EXACT solutions. So iteration will not give the required solution.
Reply 12
steve2005
You said the question required EXACT solutions. So iteration will not give the required solution.

yeah but still better than nothing right?
do you know how to find the other solution?
How about substitution?
let u=2x u=2x
so eu=5 e^u = 5
therefore u=ln5 u={\ln {5}}
use that to find x x
Reply 14
I agree with 12ln(5)\frac{1}{2}ln(5)

I did it by differentiating both sides w.r.t x and it becomes simple to solve. However, the graphical idea of there being two solutions seems quite completing.

I might have to get MatLab out.

EDIT: How does the substitution work?
Reply 15
ollimollimoll
How about substitution?
let u=2x u=2x
so eu=5 e^u = 5
therefore u=ln5 u={\ln {5}}
use that to find x x


it needs a u after the 5, e^u = 5u
desertwarrior
yeah but still better than nothing right?
do you know how to find the other solution?


This is not exact, obviously

Henerz


EDIT: How does the substitution work?


Good old substitution:
replace 2x with u, and you can work it out easily. Then, afterwards when solved with u, put 2x back in. Simple :smile:

EDIT:that doesn't work, sorry
Monochrome
it needs a u after the 5, e^u = 5u

does it?

Ah! you are right. oh well, I tried...
Reply 19
ollimollimoll
Good old substitution:
replace 2x with u, and you can work it out easily. Then, afterwards when solved with u, put 2x back in. Simple :smile:

i think that is wrong

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