Answer:
What this rule says, in practical terms, is that you can evaluate a non-standard-base log by converting it to the fraction of the form "(standard-base log of the argument) divided by (same-standard-base log of the non-standard-base)". I keep this straight by looking at the position of things. In the original log, the argument is "above" the base (since the base is subscripted), so I leave things that way when I split them up:
argument "x" in top log, base "b" in bottom log
Here's a simple example of this formula's application:
Evaluate log3(6). Round your answer to three decimal places.
The argument is 6 and the base is 3. I'll plug them into the change-of-base formula, using the natural log as my new-base log:
3(6)=ln (6) ln(3) log 3
(6)=
ln(3)
ln(6)
=
1.79175946923...
1.09861228867...
=
1.09861228867...
1.79175946923...
=
1.63092975357...
=1.63092975357...
Then the answer, rounded to three decimal places, is:
log3(6) = 1.631
I would have gotten the same final answer if I had used the common log instead of the natural log, though the numerator and denominator of the intermediate fraction would have been different from what I displayed above:
3
(
6
)
=
log
(
6
)
log
(
3
)
log
3
(6)=
log(3)
log(6)
=
0.778151250384...
0.47712125472...
=
0.47712125472...
0.778151250384...
=
1.63092975357...
=1.63092975357...
As you can see, it doesn't matter which standard-base log you use, as long as you use the same base for both the numerator and the denominator.
Step-by-step explanation:
