Thread: A logaritm problem from an old trig textbook

1. A 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

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

How do I get from the problem to the solution.

Thank you

2. Re: A logaritm problem from an old trig textbook

Originally Posted by Crisor2431
The equation is a logarithm problem that reads:

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

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

How do I get from the problem to the solution.

Thank you
$a^2 e^{-ax} = b^2 e^{-bx}$

Take the natural log of both sides:
$ln \left ( a^2 e^{-ax} \right ) = ln \left ( b^2 e^{-bx} \right )$

$ln(a^2) - ax = ln(b^2) - bx$

Can you finish from here?

-Dan

3. Re: A logaritm problem from an old trig textbook

I believe so. Thank you.

4. Re: A logaritm problem from an old trig textbook

Hello, Crisor2431!

I assume that you know that $\log_e(x)$ is written $\ln(x).$

(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: . $a^2e^{-ax} - b^2e^{-bx} \:=\:0 \quad\Rightarrow\quad b^2e^{-bx} \:=\:a^2e^{-ax}$

. . . . . . . . $\frac{e^{-bx}}{e^{-ax}} \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad e^{ax-bx} \:=\:\tfrac{a^2}{b^2} \quad\Rightarrow\quad e^{(a-b)x} \:=\:\left(\tfrac{a}{b}\right)^2$

$\text{Take logs: }\:\ln\left(e^{(a-b)x}\right) \;=\;\ln\left(\tfrac{a}{b}\right)^2 \quad\Rightarrow\quad (a-b)x\underbrace{\ln(e)}_{\text{This is 1}} \:=\:2\ln\left(\tfrac{a}{b}\right)$

Therefore: . $x \;=\;\dfrac{2\ln(\frac{a}{b})}{a-b}$