Karl's Calculus Tutor - Coached Exercise 6.3-8

Coached Exercise 6.3-8KCT logo

© 1997 by Karl Hahn

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 
                    xh - 1
   ln(x)  =   lim         
             h  > 0    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 

                     (ab)h - 1
   ln(ab)  =   lim            
              h  > 0     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 

                     ahbh - 1
   ln(ab)  =   lim           
              h  > 0     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 

   (ah - 1)(bh - 1)
which multiplies out to 
   ahbh - ah - bh + 1
We can't just go changing the numerator to this. But you can introduce the ah term by adding it and subtracting it. That is the same as adding ah - ah, which is the same as adding zero and doesn't change anything. Likewise with the bh 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 

                     ahbh - ah - bh + 1 + ah + bh - 2
   ln(ab)  =   lim                                   
              h  > 0               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 

                     (ah - 1)(bh - 1) + ah + bh - 2
   ln(ab)  =   lim                                 
              h  > 0                h

Step 4: Break the fraction in two. Separate the factored stuff from the residual terms. Write the new equation, then click here.

You should now have 

                     (ah - 1)(bh - 1)   ah + bh - 2
   ln(ab)  =   lim                    +            
              h  > 0         h               h

Step 5: Observe that the left hand fraction contains a familiar expression. Can you see it as 

   ah - 1
          (bh - 1)
      h
But in the limit, we know that 
   ah - 1
         
      h
becomes ln(a) and (bh - 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 

                     ah + bh - 2
   ln(ab)  =   lim              
              h  > 0      h
If you have a good eye for algebra, you might be able to complete the derivation from this point.

Step 6: 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 

                     ah - 1   bh - 1
   ln(ab)  =   lim          +       
              h  > 0    h        h
The rest should be a piece of cake.

Step 7: 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? 

                    xh - 1
   ln(x)  =   lim         
             h  > 0    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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