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?

log(a-b) ¹ loga-logb in general.

loga-logb º log(a/b) is an identity.

Kerwin

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

thanks

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

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?

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