Question

Find three possible rearrangements of the equation f(x)=0 into the form x=F(x).

B) \[f(x) = e^x - {5 \over x}\]

Answer 1

Iterations

Answer 2

@satellite73 @zarkon @amistre64 @mylinaya

Answer 3

Thanks .Sam.

Answer 4

as in newton ralphson?

Answer 5

oh i think i see
start with
\[e^x-\frac{5}{x}=0\]
\[e^x=\frac{5}{x}\]
\[x=5e^{-x}\]

Answer 6

are there really 3 different forms of this? let me know and i will learn something

Answer 7

as in eg.
f(x) = x^3 - 2x + 3

becomes

x= cube root of {2x - 3}
or
X^3 = 2x-3
or
(2x-3)^1/3

Answer 8

sorry, the last is \[x= (2x-3)^{1/3}\]

Answer 9

only one of those looks like
\[x=F(x)\] namely the first one

Answer 10

oh i think i get it. now i need to think

Answer 11

ok lets try taking the log

Answer 12

\[e^x=\frac{5}{x}\]
\[x=\ln(\frac{5}{x})\]

Answer 13

or if you prefer
\[x=\ln(5)-\ln(x)\] how about that one?

Answer 14

The books answers are:

\[x=5e^{-x}\]

\[x= \ln5-lnx\]

\[x=\sqrt {5xe^{-x}}\]

Answer 15

ok two out of three isn't bad...

Answer 16

No, but how did you get them?

Answer 17

i started with
\[e^x=\frac{5}{x}\] and for the first one i solved for x on the right hand side
\[e^x=\frac{5}{x}\]
\[\frac{1}{e^x}=\frac{x}{5}\]
\[\frac{5}{e^x}=x\]
\[5e^{-x}=x\]

Answer 18

secone one i started with
\[e^x=\frac{5}{x}\] and solved for the x on the left hand side
\[e^x=\frac{5}{x}\]
\[x=\ln(\frac{5}{x})\]
\[x=\ln(5)-\ln(x)\]

Answer 19

i have to think about how to get the third one

Answer 20

i have to say the third one looks artificial, by which i guess i mean too complicated. we can get there by algebra, but i am not sure why you would want to do so

Answer 21

Thanks for your help!

Answer 22

yw

Answer 23

Is it possible you help me with the new one I wrote up?