Week 9 Topic 1 – Equations with Logarithms
Week 9
Topic 1 – Equations with Logarithms
Introduction
Solving equations can be pretty fun. But these equations can be pretty hard. We’re
going to use properties of logarithms extensively, but first review the two properties of
equations you can always use:
1. You can add or subtract a term from both sides of the equation
a. Remember that terms are separated by + and – signs. So in an equation like
3ln( x) = 7 − 2 ln( x) you can add 2 ln( x) to both sides of the equation.
2. You can multiply or divide every term in the equation by a number
If your equation-solving skills are rusty, you will definitely want to do some review before
adding logarithms into the problems. I suggest www.KhanAcademy.org.
Key Points
1. Be able to solve equations with logarithms and exponents
a. Don’t forget about factoring and solving quadratics here
2. Know how to check your answers and identify extraneous solutions
Reading
This topic is mostly just practice. There are a few examples below, but also read over the
textbook on pages 334 – 337. All the examples are good to read in detail and work out for
yourself. Example 6 is not something you usually encounter, but read it through and make sure
you can follow the logic.
There are a few ways to use properties of logarithms to solve equations. Let’s look at an
example of each of those, and then do a couple more complicated examples.
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Week 9 Topic 1 – Equations with Logarithms
Property 3: Sum of Logs
Example 1: Solve 5 − 2 log( x) = 8log( x)
Solution: For this example, notice you have x in two places in the equation. Both of
them are inside logarithms. In order to solve, you’ll have to get an equation with an x in just
one place.
We can add and subtract terms from each side of the equation. Here’s the same
equation, but with boxes around every term:
5 − 2 log( x) = 8log( x)
We want all the terms with x on one side of the equation, and all the terms without x on
the other side, so let’s add 2 log( x) to both sides of the equation:
5 = 10 log( x)
Notice that 2 log( x) and 8 log( x) made 10 log( x) , just like 2y and 8y make 10y.
Size up this equation again. There’s an x in just one spot, and it’s inside a logarithm. The
only way to get a variable out of a logarithm is to rewrite it as an equation without a logarithm.
But we can only do that once we’ve isolated the logarithm itself. There are two possible steps
we could take here:
Option 1: Divide both sides by 10:
1
= log( x)
2
Option 2: Use Property 5 of logarithms:
5 = log ( x10 )
Either one of these are valid steps. It’s common in these problems to have 2 or more
correct methods to get the solution, and you’re welcome to do whichever is more comfortable
for you. I think Option 1 is simpler, so I’m going to use that.
Remember that log( x) is the same as log10 ( x) , so we can rewrite the statement:
101/ 2 = x
x ≈ 3.1623
As always, we prefer exact answers to approximate. However, my exact answer will
sometimes look different than someone who’s followed a different series of steps, so
approximate answers are easier to compare.
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Week 9 Topic 1 – Equations with Logarithms
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We’re not done yet! We always always ALWAYS want to check our solutions. You can
use a calculator to make the check easier, but it’s not necessary in this example.
5 − 2 log (101/2 ) = 8log (101/ 2 )
1
1
= 8⋅
2
2
5 −1 = 4
5− 2⋅
The interesting thing about solving equations with logarithms is that we can solve them
correctly but still come up with solutions that don’t work. A solution that doesn’t work is called
an extraneous solution. This is a second important reason to check your work, even if you never
make mistakes.
Property 4: Difference of Logs
Example 2: Solve log 2 ( x) − log 2 (1 − x) = 0
Solution: Take a moment to look this equation over as well. Again, there are x’s in two
places, so we need to combine the logarithms. They’re both on the same side of the equation,
and one is subtracted from the other. So we use the Difference of Logs property:
 x 
log 2 
=0
 1− x 
Be careful! Don’t divide the logarithms, divide the arguments inside of a single
logarithm.
Now all the x’s are in a single logarithm, so we rewrite the statement:
x
= 20
1− x
x
=1
1− x
Now multiply both sides of the equation by (1 − x ) , and solve as usual:
x = 1− x
2x = 1
1
x=
2
Now check your work:
Week 9 Topic 1 – Equations with Logarithms
4
 1 


log 2  2  = 0
1
 1− 
 2
1
 
log 2  2  = 0
1
 
2
log 2 (1) = 0
That last line is true because you can rewrite it to say 1 = 20 , which is true.
Property 5: A Constant Times a Log
Example 3: Solve 2log 5 (r ) − log5 (36) = 0
Solution: This looks extremely similar to Example 2. We have a difference of logs equal
to zero. However, we can’t use Property 4 with that 2 in front of log 5 (r ) . We have to rewrite
using Property 5:
log5 (r 2 ) − log5 (36) = 0
Now we could combine the logs as we did in Example 2 (and it would work out the same
in the end), or we can add log 5 ( 36 ) to both sides of the equation:
log5 (r 2 ) = log5 (36)
The r2 is stuck inside that logarithm, but by the One-to-One property, we know that:
r 2 = 36
r = ±6
Check:
2log5 (6) − log 5 (36) = 0
2.22656550512 − 2.22656550512 = 0
2log5 (−6) − log5 (36) = 0
The positive solution checks out, but we can’t log a negative number, so we say r = -6 is
extraneous, and the only real solution is r = 6.
Variables in Exponents
Example 4: Solve 3 = 10 x−6
Week 9 Topic 1 – Equations with Logarithms
Solution: There are no logarithms in this problem, but the variable is in the exponent.
That means we have to use a logarithm and Property 5. Notice that Property 5 works for a
logarithm of any base, so we can use base 10, base 2, base e, or anything we like. I’m going to
work it out using the natural logarithm, but try it on your own with a different base and verify
the solutions match.
ln ( 3) = ln (10 x −6 )
ln ( 3) = ( x − 6) ln (10 )
Since the exponent is x − 6 , we have to bring the entire thing in front of the logarithm.
Now you could distribute the ln (10 ) , but since it’s a factor on the left, we can divide by it:
ln(3)
= x−6
ln(10)
ln(3)
+6= x
ln(10)
6.4771 ≈ x
It’s easy to check this on a calculator.
Longer Examples
You’ll notice that for every problem, I look at it first. When solving these equations, it’s
just as important to know what you can’t do. The properties are very specific, so there are only
a few legal moves, but hundreds of illegal possibilities. So it’s not enough to look at the right
steps; when you make a mistake, ask yourself why a particular step doesn’t work.
Example 5: Solve 10 ⋅ ( 2T ) = 5T
Solution: Okay, we have variables in exponents again, so we’ll have to logarithm both
sides. Now it doesn’t matter what base we use. Since this problem has exponents with bases of
2 and 5, one of those is probably the best choice1. I’ll use base 2:
log 2 (10 ⋅ 2T ) = log 2 ( 5T )
Now we need to use Property 5 to get the variable out of the exponent. On the right
side of the equation, there’s no impediment. But on the left side, the exponent only applies to
the 2, so we need to break apart the left side using Property 3 first:
1
For checking on a calculator, the natural logarithm is the best choice
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Week 9 Topic 1 – Equations with Logarithms
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log 2 (10 ) + log 2 ( 2T ) = log 2 ( 5T )
log 2 (10 ) + T ⋅ log 2 ( 2 ) = T ⋅ log 2 ( 5 )
log 2 (10 ) + T = T ⋅ log 2 ( 5 )
Do you see what the last step was? Using base 2 allowed for that simplification. Now
we’re still not in a great position, so size up the equation again. We have two terms with the
variable in them, so let’s get them on the same side of the equation:
log 2 (10 ) = T ⋅ log 2 ( 5 ) − T
The way to get a single T in the expression is by factoring. Then we can divide by the
coefficient and be done:
log 2 (10 ) = T ⋅ ( log 2 ( 5 ) − 1)
log 2 (10 )
=T
log 2 (5) − 1
2.5129 ≈ T
To check, type that into a calculator. If you approximate T on the calculator, your
answer will only be approximately equal, but it should still be very close.
Example 6: Solve 2 log 4 ( x) = log 4 (12 − 8 x) − 1
Solution: Again, we have the variable in two places, both inside logarithms. So we’ll
need to combine the logarithms. Before we can do that, we need to use Property 5 on the left
side of the equation.
I’m going to skip a couple steps on this example. Follow along with paper and pencil,
and see if you can figure out how I get each new line.
log 4 ( x 2 ) − log 4 (12 − 8 x ) = −1
x2
1
=
12 − 8 x 4
2
x + 2x − 3 = 0
( x + 3)( x − 1) = 0
The solutions are x = -3 or x = 1. Checking them reveals that the negative solution is
extraneous, so the only real solution is x = 1.
Week 9 Topic 1 – Equations with Logarithms
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Concluding Remarks
This can be a tough section. Make it easier by realizing that there are really only a few
types of problems. Each one requires one or two properties of logarithms. Examine the
examples here and in the book to be sure you know what moves preserve equality (are legal)
and resist the temptation to do anything crazy2.
If you use Wolfram|Alpha and show step-by-step solutions3, it will convert every
logarithm into base e first. That makes understanding their solutions a bit trickier. Further, like
many software programs, Wolfram|Alpha uses “ log( x) ” to mean ln( x) . If you want it to do
log10 ( x) , you have to enter log(10, x) .
2
Crazy things: Dividing both sides by “log”, 5 becomes 5x, dividing 2 by 2 to get x, etc.
x
x
3
If you download the app, which costs $5, you get unlimited step-by-step solutions