Solution to Exercise 6.3-8

The problem was to prove

   ln(ab)  =  ln(a) + ln(b)

using only our limit definition of ln(x), which, as you probably recall is

ln(x) = lim h → 0

x^h - 1 h

Step 1: Apply the limit definition to ln(ab). click here when you have finished that.

You simply substitute ab for x in the limit equation. You would get

ln(ab) = lim h → 0

(ab)^h - 1 h

Step 2: Apply the identity for a product taken to a power. You will apply it to (ab)^h. Click here when you've done it.

You should have gotten

ln(ab) = lim h → 0

a^hb^h - 1 h

Step 3: Be clairvoyant. It's hard for anybody to explain where the vision comes from to try a step like this. The more proofs and derivations you do, the more easily the vision comes. The objective here is to end up with something in the numerator that we can factor. Specifically we want the numerator to factor into

   (a^h - 1)(b^h - 1)

which multiplies out to

   a^hb^h - a^h - b^h + 1

We can't just go changing the numerator to this. But you can introduce the a^h term by adding it and subtracting it. That is the same as adding a^h - a^h, which is the same as adding zero and doesn't change anything. Likewise with the b^h term. We can change the -1 to a +1 by adding and subtracting 2, that is adding 2 - 2 to the numerator. Go ahead and do these things and then see how you can munge it to get the factorable stuff plus some residual terms. Write that equation, then click here.

You should have gotten

ln(ab) = lim h → 0

a^hb^h - a^h - b^h + 1 + a^h + b^h - 2 h

Step 4: Factor what you can factor. We already saw that the first four terms of this numerator can be factored. Factor them and write the new equation. Click here when you're done.

You should have gotten

ln(ab) = lim h → 0

(a^h - 1)(b^h - 1) + a^h + b^h - 2 h

Step 5: Break the fraction in two. Separate the factored stuff from the residual terms. Write the new equation, then click here.

You should now have

ln(ab) = lim h → 0

(a^h - 1)(b^h - 1) a^h + b^h - 2 + h h

Step 6: Observe that the left hand fraction contains a familiar expression. Can you see it as

   a^h - 1
          (b^h - 1)
      h

But in the limit, we know that

   a^h - 1
         
      h

becomes ln(a) and (b^h - 1) becomes zero. That makes their product zero in the limit as well. So in the limit, the entire left hand fraction is zero. And that leaves you with what? Rewrite the equation without the stuff that goes to zero in the limit, then click here. But make sure you understand why this step is able to get rid of all that left hand fraction stuff first.

You should be left with

ln(ab) = lim h → 0

a^h + b^h - 2 h

If you have a good eye for algebra, you might be able to complete the derivation from this point.

Step 7: One plus one equals two. Express the -2 as two -1's. Then rearrange the terms and break the expression into the sum of two familiar looking fractions. When you are done, click here.

You should now have

ln(ab) = lim h → 0

a^h - 1 b^h - 1 + h h

The rest should be a piece of cake.

Step 8: Apply the limit formula to each of the new fractions. Remember that the limit of the sum is the sum of the limits. And what is the limit of each of those fractions according to our original limit formula?

ln(x) = lim h → 0

x^h - 1 h

Just substitute the appropriate variable in for x in each case and you get

   ln(ab)  =  ln(a) + ln(b)

which is what we were trying to prove all along.

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