I’ve been on this equation for a while. Can you help?

[Mathmasteriw]

Hi guys and girls,
Can any of you out there help me with this equation? Not sure where to go form here what to do next. Attched is the photo of the question and where ive got to.
Thanks for your time!

 

[Deleted member 4993]

Hi guys and girls,
Can any of you out there help me with this equation? Not sure where to go form here what to do next. Attched is the photo of the question and where ive got to.
Thanks for your time!

You got:

\(\displaystyle \frac{V_c}{V_s} \ = \ \ 1 \ - \ e^{\left(\frac{-t}{Rc} \right)}\)

\(\displaystyle 1 - \ \frac{V_c}{V_s} \ = \ e^{\left(\frac{-t}{Rc} \right)}\)

Now take log of both sides and continue.....

 

[Mathmasteriw]

You got:

\(\displaystyle \frac{V_c}{V_s} \ = \ \ 1 \ - \ e^{\left(\frac{-t}{Rc} \right)}\)

\(\displaystyle 1 - \ \frac{V_c}{V_s} \ = \ e^{\left(\frac{-t}{Rc} \right)}\)

Now take log of both sides and continue.....

Great ! Thank you for your help! I will try agian tomorrow!
Really appreciate you’r help!

 

[Mathmasteriw]

Hi guys and girls, I’ve been stuck on this for ages! Any help greatly appreciated.
Thanks for your time.

 

[Romsek]

Aside from not using parens to keep things clear you're doing fine so far.

What's the problem? You just have a couple trivial algebra steps left.

 

[Mathmasteriw]

N

Aside from not using parens to keep things clear you're doing fine so far.

What's the problem? You just have a couple trivial algebra steps left.

Not sure of the next steps and how to find C with regards to the 'log' in the equation

 

[Romsek]

[MATH]\log_e\left(1-\dfrac{V_c}{V_s}\right) = -\dfrac{t}{RC}[/MATH]
that's where you left off. Simple algebra from here.

[MATH]C \log_e\left(1-\dfrac{V_c}{V_s}\right) = -\dfrac{t}{R}\\~\\ C = -\dfrac{t}{R \log_e\left(1-\dfrac{V_c}{V_s}\right)}[/MATH]
I suppose I should note that this is only valid for [MATH]V_c \neq V_s,~\text{i.e. }t\neq 0[/MATH]

 

[Mathmasteriw]

[MATH]\log_e\left(1-\dfrac{V_c}{V_s}\right) = -\dfrac{t}{RC}[/MATH]
that's where you left off. Simple algebra from here.

[MATH]C \log_e\left(1-\dfrac{V_c}{V_s}\right) = -\dfrac{t}{R}\\~\\ C = -\dfrac{t}{R \log_e\left(1-\dfrac{V_c}{V_s}\right)}[/MATH]
I suppose I should note that this is only valid for [MATH]V_c \neq V_s,~\text{i.e. }t\neq 0[/MATH]

I tried nearly everything and still did not get this! amazing work thank you!
Now I am unsure what to do with the 'e'? I dont seem to have a value for e so need to look into this! but thanks so much for your help!

 

[Dr.Peterson]

Now I am unsure what to do with the 'e'? I dont seem to have a value for e so need to look into this! but thanks so much for your help!

Are you aware that [MATH]\log_e[/MATH] is commonly called [MATH]\ln[/MATH] on calculators (or in programming languages)? You don't need to do anything with e beyond that.

 

[Mathmasteriw]

Ofcourse Dr Peterson! everyday a school day thank you very much Doc! It's been a long day! thank you!!