Isolating a variable in an exponent

[thatguythere]

Homework Statement

So I had to find out
N=N_oe^-ux
Where N_o=1.5,e is the elementary charge, u=-0.068 and x=0.07
I came to the answer 1.221

Now I need to find x if u =-0.036, N=1.221 and N_o remains 1.5

Homework Equations

The Attempt at a Solution

I reduce the equation to the following

ln(N/N_o)=-ux

The problem is that when I plug my numbers from the original problem into this I end up getting the following

ln(1.221/1.5)=-(-0.068)x
-0.2058=0.068x
-3.02= x

This is clearly incorrect since I know x should be 0.07.
Any help would be greatly appreciated.
Thank you.

 

[Mentallic]

Homework Statement

So I had to find out
N=N_oe^-ux
Where N_o=1.5,e is the elementary charge, u=-0.068 and x=0.07
I came to the answer 1.221

I get something different.

Now I need to find x if u =-0.036, N=1.221 and N_o remains 1.5

I'll assume u = -0.036 is a typo because it looks as though you want to use u = -0.068 again.

Just check your answer for N again, and see if that changes anything.

 

[thatguythere]

My order of operations was wrong then?

So I get 1.273

Now -0.036 was NOT a typo, however I am ignoring it, as I am using the original numbers to try to figure out how to isolate x. However, when I plug this new number in, I still end up with x = -2.413 instead of 0.07 which it should be.
Clearly I am doing something very wrong and have no idea what.

 

[thatguythere]

Let me illustrate step by step.

N=N_oe^-ux
N=1.5*1.6x10^{-19(0.068)(0.07)}
N=1.5*0.096^0.07
N=1.5*0.849
N=1.274

Now I try to reverse it to figure out how to isolate x, this is where I do something very wrong I believe.

N=N_oe^-ux
1.274=1.5*1.6x10^{-19(0.068)(x)}
0.849=1.6x10^{-19(0.068)(x)}
ln0.849=0.068x
-0.164=0.068x
-2.41=x

But we know x=0.07. I'm lost.

 

[Mentallic]

My order of operations was wrong then?

So I get 1.273

Sorry, I mistook e as being the natural exponential 2.718...
We're supposed to be using $e=1.602\cdot10^{-19}$, right?

Using the figures:
$$N_0 = 1.5$$
$$e=1.602\cdot 10^{-19}$$
$$u=-0.068$$
$$x=0.07$$

You should get $N\approx 1.22$

 

[Mentallic]

N=1.5*1.6x10^{-19(0.068)(0.07)}
N=1.5*0.096^0.07

How did you go from the first to the second step?

 

[thatguythere]

I got it.

 

[thatguythere]

Haha Ok so I was correct in the first place when I posted 1.221. Then I was trying to figure out what I did wrong which would explain your confusion over my next process.

 

[Mentallic]

I did 1.6x10^-19(0.068)
Is that improper order of operations?

Yes, it's not correct.

$$(n\cdot a^b)^{cd} = n^{cd}\cdot a^{bcd}$$

But you've done something else.
Of course, calculating them in this way is the long and inefficient way, so you'd instead enter it as so:

(1.6E-19)^(0.068*0.07)

EDIT: most of this is redundant, you already know it

 

[Mentallic]

Haha Ok so I was correct in the first place when I posted 1.221. Then I was trying to figure out what I did wrong which would explain your confusion over my next process.

Oh yeah sorry, that's my fault!

As for the second part of the problem where you're trying to find x, you've done essentially what confused me as well.

ln(N/No)=-ux

Remember that ln = log_e is the natural logarithm, or in other words, it reverses the natural exponential operator.

If $a=e^{b}$ then $\ln(a)=b$ but we don't have the natural exponential (e=2.718...) here, we have another base altogether, so what we'd need to do to reverse the exponent operation is to take the log with base $1.6\cdot 10^{-19}$

But your calculator probably doesn't have the option of choosing any base for the logarithm, so you instead use the formula

$$\log_ab=\frac{\log_cb}{\log_ca}$$

So we have that

$$\log_{1.6\cdot 10^{-19}}(1.221/1.5) = \frac{\ln (1.221/1.5)}{\ln(1.6\cdot 10^{-19})}$$

 

[thatguythere]

Let me try then.

 

[thatguythere]

Are you saying that
$\frac{ln(1.221/1.5)}{ln(1.6E-19)}$=-ux?

 

[Mentallic]

Yep!

 

[thatguythere]

Yeah, I got that. So using -0.036, I come up with 0.132. Excellent.

 

[thatguythere]

Thank you very much.

 

[Mentallic]

You're welcome! And sorry about the confusion at the start.