Solve the following equatiom

[eddievic]

Homework Statement

In the formula theta = Ve^(-Rt/L), the value of theta = 58, V = 255, R = 0.1 and L = 0.5. Find the corresponding value of t.

Homework Equations

The Attempt at a Solution

Rt/L = V - θ



t = (L/R)(V - θ)



t = (0.5/0.1)(255 - 58) = 985



t = 985

anywhere near?

 

[DrClaude]

Rt/L = V - θ

Where did that come from?

 

[eddievic]

Where did that come from?

think I did that incorrectly should it be:
θ = Ve-Rt/ L

Where, θ = 58, V = 255, R = 0.1, and L = 0.5.

58 = 255 e-.01 t/0.5

e-t/5 = 4.3966 (approx)

Taking log on both sides

-t/5 = ln(4.3966)

-t = 7.40415

t = -7.40415

 

[DrClaude]

e-t/5 = 4.3966 (approx)

That line is not correct.

 

[eddievic]

That line is not correct.

I think at that point I should divide by 255

so

e^0.1t/0.5=0.2274

then turn into logarithmic statement

0.1t/0.5=In0.2274

then
0.1t=0.5In0.2274

then
t=0.5In0.2274/0.1

right track?

 

[DrClaude]

I think at that point I should divide by 255

so

e^0.1t/0.5=0.2274

then turn into logarithmic statement

0.1t/0.5=In0.2274

then
0.1t=0.5In0.2274

then
t=0.5In0.2274/0.1

right track?

Be careful with the minus signs. And pay attention to your notation, as that last line looks like
$$
t = 0.5 \ln \frac{0.2274}{0.1}
$$

 

[eddievic]

Be careful with the minus signs. And pay attention to your notation, as that last line looks like
$$
t = 0.5 \ln \frac{0.2274}{0.1}
$$

as a point how would I correctly write that last part to avoid confusion?

and after that I get

t=0.5In0.2274
/0.1

so t =-7.408 ?

 

[DrClaude]

as a point how would I correctly write that last part to avoid confusion?

and after that I get

t=0.5In0.2274
/0.1

If you don't want to use LaTeX notation, try

t = (0.5/0.1) ln 0.2274

so t =-7.408 ?

The numerical value is correct, but the sign is wrong.

 

[eddievic]

If you don't want to use LaTeX notation, try

t = (0.5/0.1) ln 0.2274


The numerical value is correct, but the sign is wrong.

I think I made a typo

t = 7.405

for future reference is there a 'sticky' guide to using LaTeX notation as I'm currently doing a course and am finding Physics Forums to be of some use!

 

[DrClaude]

Check out https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[eddievic]

Check out https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

Great thanks for all you help!

 

[Mark44]

Homework Statement

In the formula theta = Ve^(-Rt/L), the value of theta = 58, V = 255, R = 0.1 and L = 0.5. Find the corresponding value of t.

Homework Equations

The Attempt at a Solution

Rt/L = V - θ

No.

t = (L/R)(V - θ)

t = (0.5/0.1)(255 - 58) = 985

t = 985

anywhere near?

The first two things you should have done, in this order, are:
1. Divide both sides of the equation by V.
2. Take the natural log of both sides.

Starting with θ = Ve^(-Rt/L), step 1 yields
θ/V = e^(-Rt/L)

Taking logs of both sides:
ln(θ/V) = -Rt/L

Now solve for t.

 

[Ridley01]

Hi,

I have the same question in an assignment but get a different answer can anyone advise where I am going wrong please here is my working out.

θ=Ve^(-Rt)/L

58 = 255e^(-0.1t)/0.5

58/255 = e^(-0.1/t)/0.5

0.2275 = e^(0.1/t)/0.5

(-0.1t)/0.5 = Log(0.2275)

-0.1t = 0.5 Log(0.2275)

t = (0.5Log(0.2275))/-0.1

t = -0.3215/-0.1

t = 3.215

 

[xiavatar]

@Ridley Remember that your taking the natural log, and not log base 10. So

$-.1t=.5ln(.2275)$.

 

[Ridley01]

@Ridley Remember that your taking the natural log, and not log base 10. So

$-.1t=.5ln(.2275)$.

So I should have wrote.


-0.1t = 0.5ln(0.2275)

t = (0.5ln(0.2275))/-0.1

t = -0.7403 / -0.1

t = 7.403

 

[xiavatar]

Yes, that is correct. But do you understand the difference? When you take the logarithm of an exponential you can only "cancel" out the base if and only if the base of your exponent is the same as the base of your logarithm. In other words

$log_a(a^n)=n*(log_a(a))=n*(1)=n$

It is not true however when $a\neq b$ as in the example below

$log_b(a^n)=n*(log_b(a))\neq n$