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.
