Expert: Josh - 9/18/2009

Question
Hi there, I have just been handed a piece of work for college and one of the questions is really irritating me...
It says...
Using your Scientific calculator, work out the following:

The problem is...   8 log (10) 2.5            the '(10)' is in lower case form...

If you could answer this question for me it would be much appreciated!

Regards, Lew

Answer
Hi Lew,

I can take a guess, but the exact procedure varies depending on the calculator you use. Some calculators require the user to enter all the numbers and operations in the same order as they appear on paper. In this case, you will hit the following buttons: [8] [*] [log_10] then a left parenthesis may appear, this should be followed by [2] [.] [5] and a closing right parenthesis [)] if necessary. Finally, you press the [ANS] or [=] button.

For other models, the log_10(2.5) part is computed as follows: you supply the argument that the log function takes, in this case, this would be [2] [.] [5], then you press the [log_10] button and the answer emerges. After this, you can multiply this figure by 8 to obtain the final answer.

In terms of the meaning of the question, log_10(2.5) returns an exponent (say, y) that when 10 is raised to the power of y, it yields 2.5. In other words, if y=log_10(2.5), then 10^(y) = 2.5.

A few more comments regarding logarithms.

CHANGE OF BASE RULE
In general, log_a(x)=log_b(x)/log_b(a). So, if you have to compute log_e(3) for example, where log_e is the natural logarithm (i.e., the base "e" is a numerical constant with an approximate value of 2.718281), and there isn't a "log_e" or "ln" button on your calculator, you can instead compute log_10(3)/log_10(2.718281...), for instance, to get the answer. In fact, b can be any real positive number. So, you may also use log_2(3)/log_2(2.718281...) to calculate log_e(3).
As another example, we can write log_7(2) = log_10(2)/log_10(7). The second base "b" is chosen usually for convenience.

HOW LOG WORKS

c*log_a(b) = log_a(b^c).

e.g., 8*log_10(2.5) = log_10(2.5^8)   Try it and see.

Regards,
Josh