Solving 3^x - 2^x = a

By Anonymous on Monday, April 30, 2001 - 03:44 pm :

3^x - 2^x = a

I have tried to solve this equation as follows:

ln 3^x - ln 2^x = ln a
x ln 3 - x ln 2 = ln a
x(ln(3/2)) = ln a
x = (ln a)/(ln(3/2))

But this does not seem to work. If x=2, then a should be 5. Maybe the error is caused by dividing by the unknown required quantity(x) which leads to the loss of some solutions. Can anyone help here?

By Kerwin Hui (Kwkh2) on Monday, April 30, 2001 - 05:45 pm :

Anonymous,

$\log (a - b) \neq \log a - \log b$ in general.

$\log a - \log b \equiv \log (a / b)$ is an identity.

Kerwin

By Anonymous on Thursday, May 3, 2001 - 04:00 pm :

But if the log(a/b) is identical with log a - log b, why does it fail as a component in this equation?

thanks

By Kerwin Hui (Kwkh2) on Thursday, May 3, 2001 - 04:51 pm :

The first mistake comes in the 2nd line, when you say

ln (3^x - 2^x ) = ln 3^x - ln 2^x = ln a

Kerwin

By Anonymous on Sunday, May 6, 2001 - 12:24 pm :

Ok, I can see the mistake now. But if I can't separate ln(3^x - 2^x ) how do I go about solving this?

By Kerwin Hui (Kwkh2) on Sunday, May 6, 2001 - 03:10 pm :

The equation belongs to a class of equation that is known to have no way of solving it exactly. The best you can do is to ask a computer for a numerical answer.

Kerwin