For the following example problems the instruction is to solve for x
algebraically.
Example 1:
Solution:
Explanation:
Here all of the numbers in the problem are powers of 3, so even though
log base 3 is not on the calculator we can solve the problem by taking
logs base 3 on both sides, because we can find the logs base 3 of all
of the numbers involved by thinking about it. The left side of line 2
comes from one of the
inverse properties of
logarithms. I have colored the expression in the exponent to make
it easier to see as representing a single number, because sometimes it
is difficult to see that anything you can do with x you can do with
2x-1. For the rest of the solution we just solve it like any other
linear equation, adding 1 to both
sides and then dividing both sides by 2.
Example 2:
Solution:
Explanation:
The numbers in this problem are not as nice as those in Example 1,
because there is no number that 5 and 13 are nice powers of. In this
kind of problem it is best to take log on both sides with a base that
the calculator can do, either the common log or the natural log. I have
chosen here to use the common log, the log base 10, but it would have
worked just as well to have use the natural log, the log base e. Then
instead of using an
inverse property of logs we
use the
property about logs of powers.
Again I have colored the 2x in hopes that you can more easily see it as
behaving just like a single variable. Anything you can do with x you
can do with 2x, because if x is a real number 2x is a real number too.
In the 3rd line I rearranged things to make the 2log5 look more like a
proper coefficient. Then in the next step we divide both sides by it
and then get out the calculator to get a 4 place decimal approximation
to the answer.
Example 3:
Solution:
Explanation:
With e in the problem it is best to use the natural log ln, since ln
means log base e. First we divide both sides by 6 to get the
exponential alone. In the last line I used a little trick in order to
save calculator keystrokes. It was not entirely necessary. It would
also be fine to just key in .5 in the calculator and then write down
the opposite of what you get.
Example 4:
Solution:
Explanation:
With a logarithmic equation usually the best way to solve it is to
exponentiate both sides, that is turn both sides into exponents for the
base of the logarithm, in this case 3. Note that this is not at all the
same thing as cubing both sides. Then we get the left side of the
second line from one of the
inverse properties
and the right side is simply raising 3 to the second power, that is
multiplying it by itself.
Example 5:
Solution:
Explanation:
In this problem, before we can get rid of the logs we need to use the
log properties to write the left side a
single logarithm. It wouldn't do to just raise 2 to each term. You can
only do that with multiplying both sides of an equation by something
because of the distributive property, and there is no distributive
power of exponentiating. So what we have to do is use the
product property backwards to get the left
side of our first line. After that we can exponentiate both sides with
a base of 2 to get rid of the log base 2. After that it becomes a
quadratic equation that we can
solve by factoring. But the -2 solution has to be thrown out, because
in the original equation it would require taking a log of a negative
number, which is not allowed.
Example 6:
Solution:
Explanation:
When you have a problem with the log of a log, you just have to undo
the logs one at a time. First we raise 2 to both sides to get rid of
the log base 2, and then we raise 3 to both sides to get rid of the log
base 3 and get our final answer.