Steps in an Equation

I was reading about ordinary differential equations.

The following lines appear in an equation in the book: ln=natural log

-(L/R)ln(V-Ri) = t - (L/R)lnV

(L/R)ln(V/(V-Ri)) = t

V/(V-Ri) = e (to the power of Rt/L)

Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

Thank you! :smile:

Reply 1

Prime Suspect

Original post by little pixie

I was reading about ordinary differential equations.

The following lines appear in an equation in the book: ln=natural log

-(L/R)ln(V-Ri) = t - (L/R)lnV

(L/R)ln(V/(V-Ri)) = t

V/(V-Ri) = e (to the power of Rt/L)

Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

Thank you! :smile:

\dfrac{L}{R}\ln(\frac{V}{V-R_i}) = t

Multiply both sides by R, divide by L:

\ln(\frac{V}{V-R_i}) = \dfrac{Rt}{L}

Take exponentials of both sides

\dfrac{V}{V-R_i} = e^{\frac{Rt}{L}}

EDIT

and for step 1-2:

\dfrac{-L}{R}\ln({V-R_i}) = t - \dfrac{L}{R}\ln(V)

so

\dfrac{-L}{R}\ln({V-R_i}) + \dfrac{L}{R}\ln(V) = t

then factorise L/R

\dfrac{L}{R}[\ln(V) - \ln({V-R_i})] = t

use that

\ln(a) - \ln(b) = \ln(\frac{a}{b})

to get

\dfrac{L}{R}[\ln(\frac{V}{V-R_i})] = t

(edited 13 years ago)

Reply 2

jaheen22

12

From step 1 to 2. The log term is being brought from the right hand side to the left and a common factor of (L/R) is being take out of both log terms. Then by using the laws of logs: ln A - ln B = ln (A/B).

From step 2 to 3 (L/R) is being brought to the right hand side of the equation and then both sides are being raised as powers of e.

I hope you can follow this :s-smilie:

Reply 3

Farhan.Hanif93

18

Original post by little pixie

I was reading about ordinary differential equations.

The following lines appear in an equation in the book: ln=natural log

-(L/R)ln(V-Ri) = t - (L/R)lnV

(L/R)ln(V/(V-Ri)) = t

V/(V-Ri) = e (to the power of Rt/L)

Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

Thank you! :smile:

From 1 to 2, they have added

\frac{L}{R}\ln V

to both sides and recalled the law of logs which states that

a\ln (b) - a\ln c = a\ln (\frac{b}{c})

for use on the new LHS following the addition I mentioned.

From 2 to 3, they rearrange first for

\ln \frac{V}{V-Ri}

and then note that

e^{\ln x} = x

so they raised both sides as a power of e to get rid of the log.

Reply 4

little pixie

OP

12

Thanks very much guys. That's a great help. It looks a lot easier now once explained!

Reply 5

Vib-Rib

7

Re write

-(\frac{L}{R})ln(V-Ri)=t-(\frac{L}{R})lnV

as

-(\frac{L}{R})ln(V-Ri)+(\frac{L}{R})lnV=t

Factor out the -(L/R) on the left side to get

-(\frac{L}{R})(ln(V-Ri)-lnV)=t

when you take the log of one thing away from the log of another thing it's the same as taking the log of their quotient. so....

-(\frac{L}{R})ln(\frac{V-Ri}{V})=t

getting closer now, lets add -(L/R) to both sides of the equation and raise e to the power of both sides of the equation the get rid of the natural logs.

ln(\frac{V-Ri}{V})=t+\frac{L}{R}

\frac{V-Ri}{V}=e^{t+\frac{L}{R}}

Hope this helps you understand....Damn I hate LaTeX.

(edited 13 years ago)

Reply 6

Vib-Rib

7

Original post by little pixie

I was reading about ordinary differential equations.

The following lines appear in an equation in the book: ln=natural log

-(L/R)ln(V-Ri) = t - (L/R)lnV

(L/R)ln(V/(V-Ri)) = t

V/(V-Ri) = e (to the power of Rt/L)

Can anyone explain the steps taken going from line 1 to 2 and 2 to 3.

If there's any missing lines, could you let me know. I can't seem to figure out how they've taken these steps.

Thank you! :smile:

[edit]sorry double post[/edit]

Steps in an Equation

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