Section 4.5B: Using Log Laws to Solve Equations
In Sections 4.2 and 4.3, we solved some exponential equations by the following main steps:
1. Use a common base (a small number greater than 1, usually).
2. Use laws to combine the exponentials.
3. Cancel the bases to get a simpler equation.
We’ll do very similar steps with logarithmic equations: we collect logs with the same base together and
combine them with the laws. When we have one logarithm left on a side, we can cancel it (i.e. write exponential
form) and get a new equation type.
Warning: You need a single logarithm on any side before you can cancel. For instance, ln(x) = 3 cancels to
x = e3 , but ln(x) + ln(2) = 3 does not cancel to x + 2 = e3 ; why not?
Domain note: Logarithms are only defined for positive numbers. When you’re done, check your answers
back in the original problem. Any x value that makes a logarithm undefined is an extraneous solution; we
get rid of it from our answer!
Ex 1: Solve the following equations for x.
(a) 2 log4 (x) = 3 log4 (11)
(b) ln(x) − ln(5 − x) = 8
Note: After canceling the log, you get an equation with fractions. We multiply our equation by the
common denominator!
(c) log3 (x + 6) + log3 (x + 4) = 1
Note: Only one of the two solutions you find actually works!
(d) log2 (x + 4) = log2 (x − 4) + log3 (9) + 72 log7 (3)
Hint: We start by subtracting log2 (x − 4) to get it on the left, since the log laws can only combine logs
of the same base. You should be able to simplify the other logs, though.
Ex 2: If p denotes the selling price (in dollars) of a commodity and x is the corresponding demand (in
number sold per day), then the relationship between p and x is sometimes given by p = p0 e(−ax) , where p0 and
a are positive constants. Express x as a function of p.
Note: The answer should have p0 and a in it too... the question says “as a function of p” because p is the
only non-constant variable in the answer.
Section 4.6: Change of Base and Trickier Exponential Equations
Before now, we solved problems with several exponentials by using a common base. However, there are other
options, especially when it’s not convenient to convert everything to the same base.
Here’s the key idea: an exponential equation can be solved with any logarithm you want. If your
logarithm doesn’t cancel the base, you can still use the power law of logarithms to simplify the result!
Ex 3: Solve the equation 10x = 50 using (a) common logarithms, (b) natural logarithms, and (c) logarithms
of base 2.
Consider the forms of the three answers above. For example, you should see that log10 (50) is the same as
ln(50)/ ln(10). More generally, we can convert logs of any base to any other by using this result...
Change of Base Formula: logb (x) =
loga (x)
loga (b)
In other words, to compute a base b log in terms of base a, you find two base a logs: one of x, and the other
of your new base b. This is especially common with a = 10 or a = e, since those are the logarithms that your
calculator can compute easily.
Ex 4: Use the Change of Base formula to simplify
log(16)
log(2)
without a calculator.
Equations with Multiple Bases
The Change of Base Formula came from the idea that you could use any logarithm you wanted to solve an
equation, even if it doesn’t cancel the base. We use this same idea to handle some complicated exponential
equations. Pick any logarithm you like, use that same logarithm on each side of your equation, then simplify
with the power laws. After that, collect unknowns together.
Ex 5: Use natural logarithms to solve the equation 33x+12 = 73−9x .
Note: We can’t use the methods of Sections 4.2 and 4.3 here, since 7 is not a simple power of 3.
Question: How would the answer change if you used common logarithms instead?
Quadratics in g(x) Instead of x
When you take a quadratic expression and substitute a function for its variable, to get something like
a(g(x))2 + b(g(x)) + c
the result is a quadratic expression in g(x). Technically, it’s really a composite f (g(x)) with outer layer f (y) =
ay 2 + by + c and inner layer y = g(x).
Equation-solving strategy: Work one layer at a time! For instance, to solve a quadratic equation:
√
−b ± b2 − 4ac
2
ax + bx + c = 0
=⇒
x=
2a
When we substitute g(x), we obtain
2
a(g(x)) + b(g(x)) + c = 0
=⇒
g(x) =
−b ±
√
b2 − 4ac
2a
Thus, we solve the “quadratic outer layer” first to get values for g(x) instead of x. Next, finish solving g(x) for
x in each of your two g(x) solutions!
Sample: (x2 )2 − 3(x2 ) + 2 = 0 is quadratic in x2 . This factors as ((x2 ) − 1)((x2 ) − 2) = 0, so x2 = 1 or
√
√
x2 = 2. Solve each of these, and you get four answers x = ± 1, ± 2 .
Alternative notation: Use a new variable like y to stand for the inner function g(x). Solve ay 2 + by + c = 0
first. Once you know y’s values, you then substitute g(x) back and solve for x.
Sample revisited: y 2 − 3y + 2 = 0 leads to y = 1, 2. If y = x2 , we substitute to get x2 = 1 or x2 = 2, as
before.
Ex 6: Solve the following equations for x.
(a) x4 + x2 − 2 = 0
Hint: Since x4 = (x2 )2 , this is quadratic in x2 .
(b) e2x − 9ex = −77/4
Hint: Since e2x = (ex )2 , this is quadratic in ex . Be warned that 9ex is not (9e)x .
1
= 0.
(c) 22x − 21 · 2x + 16
Hint: This is quadratic in 2x . You should get only one solution.
Ex 7: Solve the following equation for x:
log(x1/2 ) =
p
log x
Hint: Start with the power property to pull 1/2 out of the left side. Next, how do we get rid of the square
root in the equation?
To be continued next class...
An Odd Simplification Problem (go over this outside of class)
Simplify the following completely: (e3x + e−3x )2 − (e3x − e−3x )2 .
Main idea: We use an old algebraic idea we saw before (such as with difference quotients): if we expand first,
then hopefully we can simplify like terms! Here’s a suggested order of steps:
1. Expand the squares. Remember that (e3x )2 = e6x thanks to the repeated-power property.
(e6x + 2e3x e−3x + e−6x ) − (e6x − 2e3x e−3x + e−6x )
2. Simplify e3x e−3x . One way to do this is to use the sum property and write it as e3x−3x = e0 = 1. The
other way is to note that e−3x is the reciprocal of e3x (that’s what negative exponents are!).
(e6x + 2 · 1 + e−6x ) − (e6x − 2 · 1 + e−6x )
3. Distribute the minus in the middle.
e6x + 2 + e−6x − e6x + 2 − e−6x
After the cancellation, all that’s left is 4 .