Natural logs...?

Question

The equation I have is 10e^(28/n) = e^(32/n) + 9 . I know I have to use natural logs to solve for n; I just forget the rules in how to go about this. Please help :)

Answer 1

Natural logs...? They burn cleaner than man-made logs.

You're on your own on the second question with the numbers and logarithmic scales.

Answer 2

Did you mean???

10e^(28/n) = e^[ (32/n) +9 ]


Never mind, I see how to solve the one you posted now.

Divide through by e^(28/n)

This yields:

e^(4/n) + 9 e^(-28/n) = 10

e^(4/n) + 9 [e^(4/n)] ^(-7) = 10

Let y = e^(4/n)

y + 9 / y^7 = 10

This is only true if y = 1

Thus e^(4/n) = 1

Take natural log of both sides

4/n = 0

n = positive infinity or negative infinity

Answer 3

You have to use that (a^b)^c= a^bc

So e^(28/n)= (e^4/n)^7 and (e^32/n)= (e^4/n)^8

calling e^4/n=z you get

z^8 -10z^7 +9 = 0. z=1 is a root. So trying to factorize

=(z-1)(z^7-9z6-9z^5-9z^4-9z^3-9z^2-9z-9)

I can´t find the roots of the second factor and z=1 implies

e^4/n=1 which means 4/n=0 Nonsense) Sorry