The equation is a logarithm problem that reads:

(a^2)(e^-ax)-(b^2)(e^-bx)=0 where a>b

The answer is:

(2(log[base e]a-log[base e]b))/(a-b)

How do I get from the problem to the solution.

Thank you

Printable View

- December 10th 2012, 06:44 PMCrisor2431A logaritm problem from an old trig textbook
The equation is a logarithm problem that reads:

(a^2)(e^-ax)-(b^2)(e^-bx)=0 where a>b

The answer is:

(2(log[base e]a-log[base e]b))/(a-b)

How do I get from the problem to the solution.

Thank you - December 10th 2012, 06:55 PMtopsquarkRe: A logaritm problem from an old trig textbook
- December 10th 2012, 07:13 PMCrisor2431Re: A logaritm problem from an old trig textbook
I believe so. Thank you.

- December 10th 2012, 08:12 PMSorobanRe: A logaritm problem from an old trig textbook
Hello, Crisor2431!

I assume that you know that is written

Quote:

(a^2)(e^-ax)-(b^2)(e^-bx)=0 where a>b

The answer is: (2(log[base e]a - log[base e]b)) / (a - b)

How do I get from the problem to the solution?

We have: .

. . . . . . . .

Therefore: .