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NATURAL LOGARITHMS  
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NATURAL LOGARITHMS ...

Solve the equation for x:

Problem 1: Natural Logarithms (Feb. 15, 2008) 


Solution:

Clearly the slide rule cannot be used to obtain a direct solution for this problem in its current form:

Solution 

But slide rules were never intended to do so, you needed to have some mathematical background to perform the relevant steps prior to solution.

Method 1:

Let u=ln(2x-9) and make the substitution, our equation now becomes:

Solution 

The presence of the radical implies a positive value (by convention) and a solution for u can be found by inspection; i.e. u=2, hence

ln(2x-9) = 2

We can now set the slide to determine a value for 2x-9.  If your rule has a Ln scale as on some Pickett models, it is a simple task to determine the value that has a natural logarithm of 2.

  • Move the cursor to Ln2
  • Read the result C7.4
Solution 

If not ...

  • Align C1 with LL3e
  • Move the cursor to C2
  • Read the result LL37.4

Some rules like the Faber-Castell 2/83N have a D scale synchronized with the LL3, so this is a simple task and doesn't require the slider.

Solution 

We now know that 2x-9=7.4 from which we can easily find that x=8.2 (approximate values).

 

Method 2:

This method is very similar (almost identical) to the previous one.  The only difference is the number of steps you take towards the solution.

We still make the substitution, u=ln(2x-9) so that we have:

Solution  

But this time we eliminate the radical.

Solution 

From this, we obtain two solutions, u=-3 and u=2.

We have already solved for the case u=2 and we can actually find a solution for the case u=-3, so let's do that.

  • Move the cursor to D3
  • Read the result LL030.049
Solution 

We now solve 2x-9=0.049 giving x=4.52 (approximately).

Okay, so what's the go of it?  Why has one method given only one solution and the other given two?

As some of you may known, even though the algebra is correct, we must always check that our solutions are valid.  Substituting this second solution into the original equation we achieve a negative value on the left and a positive value on the right (by convention, the radical means a positive solution only).  Hence this solution, although algebraically correct, is not a solution to the original equation. (You will notice that to solve the initial equation, we have altered its form by squaring.  Although valid, we now have a different equation.)

So the only solution to this problem is that x=8.2, always pays to check!

Just as an aside; you can get into some good conversation as to why the negative value is disregarded.  It's no point arguing that electronic calculators and CAS only give one solution.  Why?  Because they're already programmed to do it!

Enjoy!