# Convergence in logarithms

1. Sep 24, 2010

### GhostSpirit

Hi,

If I have ln{a} converging to ln{b}, can I claim that a converges to b?

I would think yes, because ln is a monotonic function and we are always sure to get a unique value of ln for any argument 'a'. Is that reasoning correct?

I am not able to prove it by the definition of convergence,

By convergence in logarithm we get: ln(a) - ln(b) < \epsilon,

ln{a/b} < \epsilon

a/b < exp{\epsilon}

(a-b)/b < exp{\epsilon} - 1

a-b < b(exp{\epsilon} - 1)

From here I can claim that since \epsilon is near to zero exp{\epsilon} is near to 1 and hence exp{\epsilon} - 1 < \epsilon_1

Thus,

a - b <b \epsilon_1.

Can I claim from here and a converges to b? The presence of b on the right hand side bothers me. Any help will be greatly appreciated.

Thanks a lot...

2. Sep 24, 2010

### mathman

You should work with absolute values. As far as the multiplication by b, it doesn't affect the analysis as long as |b| > 0.

3. Sep 25, 2010

### GhostSpirit

Thanks for the reply. I forgot to mention that a,b >0. So, as long as b is bounded, the analysis should work, right?

4. Sep 25, 2010

### Office_Shredder

Staff Emeritus
If you want to use a slightly "higher level" argument:

If ln(an) converges to ln(a), then since exponentiation is continuous, $$e^{ln(a_n)}$$ converges to $$e^{ln(a)}$$