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.
