[Guest]

can any one solve this for me

e^(3-20x)=0.4x

for x

Edited January 8, 2011 by sudhansu

[Guest]

0,2627

[Guest]

e^(3-20x)=0.4x

ln(e^(3-20x)) = ln(0.4x)

3-20x [+20x] = -0.92x [+20x]

3 [divide by 19.08] = 19.08x [divide by 19.08]

1.57=x

I hope that was right

[Guest]

e^(3-20x)=0.4x

ln(e^(3-20x)) = ln(0.4x)

ln(0.4x) does not equal x ln(0.4).

It equals ln(x) + ln(0.4)

e(3-20x)=0.4x

ln(e(3-20x)) = ln(0.4x)

3-20x [+20x] = ln(0.4x) + [+20x]

3 = ln(0.4) + ln(x) + 20x

3 [-ln(0.4)] = ln(0.4) + ln(x) + 20x [-ln(0.4)]

3 - ln(0.4) = ln(x) + 20x

3 + ln(0.4^-1) = ln(x) + 20x

3 + ln(2.5) = ln(x) + 20x

e^(3+ln(2.5)) = e^(ln(x)+20x)

[e^3] [e^(ln(2.5))] = [e^(ln(x))] [e^(20x)]

2.5 e^3 = xe^(20x)

Let assume y = 20x

Therefore

2.5 e^3 = 1/20 * ye^(y)

20 * (2.5 e^3) = 20 * 1/20 * ye^(y)

50e^3 = ye^y

There is a special function known as the Lambert W function

(http://en.wikipedia.org/wiki/Lambert%27s_W_function) that is the inverse of xe^(x)

Therefore

W(50e^3) = W(ye^y)

W(50e^3) = y

substitute 20x for y

W(50e^3) = 20x

x = 1/20 * W(50e^3)

or approximately 0.2627

Edited January 8, 2011 by spikepmw

[Guest]

e^(3-20x)=0.4x

ln(e^(3-20x)) = ln(0.4x)

ln(0.4x) does not equal x ln(0.4).

It equals ln(x) + ln(0.4)

e^(3-20x)=0.4x

ln(e^(3-20x)) = ln(0.4x)

3-20x [+20x] = ln(0.4x) + [+20x]

3 = ln(0.4) + ln(x) + 20x

3 [-ln(0.4)] = ln(0.4) + ln(x) + 20x [-ln(0.4)]

3 - ln(0.4) = ln(x) + 20x

3 + ln(0.4^-1) = ln(x) + 20x

3 + ln(2.5) = ln(x) + 20x

e^(3+ln(2.5)) = e^(ln(x)+20x)

[e^3] [e^(ln(2.5))] = [e^(ln(x))] [e^(20x)]

2.5 e^3 = xe^(20x)

Let assume y = 20x

Therefore

2.5 e^3 = 1/20 * ye^(y)

20 * (2.5 e^3) = 20 * 1/20 * ye^(y)

50e^3 = ye^y

There is a special function known as the Lambert W function

(http://en.wikipedia.org/wiki/Lambert%27s_W_function) that is the inverse of xe^(x)

Therefore

W(50e^3) = W(ye^y)

W(50e^3) = y

substitute 20x for y

W(50e^3) = 20x

x = 1/20 * W(50e^3)

or approximately 0.2627

Edited January 8, 2011 by spikepmw

[dark_magician_92]

can any one solve this for me

e^(3-20x)=0.4x

for x

taking log on both sides,

3-20x = ln(0.4) + ln(x)

3-20x=(-0.92)+lnx

lnx+20x=3.92

now this fn lnx+20x is increasing in nature, so at x=0.2,

lnx+20x=2.4

at x=0.25

lnx+20x=3.62

so it will be abt x=0.27 i think

u can make a c++ program to find exact value

all d best

[Guest]

e^(2-30x)=0.4x

(3-20x)loge^e=0.4x

3-20x = 0.4x / Loge^e

3-20x= 0.4x / 0.43

3-20x = 0.93x

3=20.93x

x=3/20.93

x= 0.143

[Guest]

http://www.numberempire.com/equationsolver.php

It's like magic!

Edit: My bad, it seems that the solver does not work for that specific equation

Edited January 10, 2011 by i9000