How do I mate '-t' the subject of the formula: v=Ve(superscript)-t/RC?
The formula is about the discharging of a capacitor and I need to find out how much time needs to have passed before the voltage across the capacitor falls to a specific level.
Thanks,
SCurley.

Is that it?

V_C=V_Oe^{-\frac{t}{RC}}

Put ln on both sides;

ln(V_C)=ln(V_Oe^{-\frac{t}{RC}})

ln(V_C)=ln(V_O) + ln(e^{-\frac{t}{RC}})

ln(V_C)=ln(V_O) - \frac{t}{RC}ln(e)

Can you continue now?

Hope it helped! :)

Is that it?

V_C=V_Oe^{\frac{-t}{RC}}

Put ln on both sides;

ln(V_C)=ln(V_Oe^{\frac{-t}{RC}})

ln(V_C)=ln(V_O) + ln(e^{\frac{-t}{RC}})

ln(V_C)=ln(V_O) + \frac{-t}{RC}ln(e)

Can you continue now?

Hope it helped! :)

Thanks for the help, but I was expecting an answer starting with -t =, or a lesson in how to transpose a formula that has a fraction to the power in it.
Could you simplify your answer, maybe that will help as it looks somewhat baffling to me.
Thanks,
SCurley

These are already very simple (note that we are now supposed to give you the answers directly);

ln(V_C) = ln(V_O)-{\frac{t}{RC}}ln(e)

ln(V_C) - ln(V_O) = -{\frac{t}{RC}}ln(e)

Since ln (e) = 1;

ln(V_C) - ln(V_O) = -{\frac{t}{RC}}

Multiply both sides by RC;

RC(ln(V_C) - ln(V_O)) = -t

These are already very simple (note that we are now supposed to give you the answers directly);

ln(V_C) = ln(V_O)-{\frac{t}{RC}}ln(e)

ln(V_C) - ln(V_O) = -{\frac{t}{RC}}ln(e)

Since ln (e) = 1;

ln(V_C) - ln(V_O) = -{\frac{t}{RC}}

Multiply both sides by RC;

RC(ln(V_C) - ln(V_O)) = -t

OK. Thank you.
SCurley.

k=3n+2
n+1 solve for n

Hi ali1986! Welcome to AMHD! :)

You should have created a thread of your own. Next time, do it, OK?

For your question, you cannot solve for n, but make n the subject of formula;

k = \frac{3n+2}{n+1}


k(n+1) = 3n+2

kn+k = 3n+2

kn-3n = 2-k

n(k-3) = 2-k

n = \frac{k-3}{2-k}

Hope that's what you were looking for! :)