Section 4.5B: Using Log Laws to Solve Equations
Recall that in Sections 4.2 and 4.3, we solved some equations with exponentials by using our laws to join
together all the exponentials with one base. We’ll do something similar with log equations.
The most common steps:
1. Bring all logarithms with the unknown on one side.
2. Use log laws to condense our logarithms into one.
3. Get rid of the log (using exponential form, or by canceling with an exponential).
4. CHECK answers in the original equation.
IMPORTANT: Remember that logarithms are only defined for positive inputs. Thus, you have to throw out
the options that make the logs undefined... those are extraneous solutions.
Ex 1: Solve the following equations for x.
(a) 2 log4 (x) = 3 log4 (11)
(b) ln(x) − ln(5 − x) = 8
NOTE: Once you get x/(5 − x) = e8 , remember that you can multiply both sides by 3 − x to get rid of
fractions! After that, collect the x’s together.
(c) log3 (x + 6) + log3 (x + 4) = 1
NOTE: This produces a quadratic equation (x + 6)(x + 4) = 31 = 3. It has two solutions, but only one
works in the original equation!
(d) log2 (x + 4) = log2 (x − 4) + log3 (9) + 72 log7 (3)
NOTE: The log laws only work for combining logs of the SAME base. The other logarithms can be
simplified away.
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 will need to have p0 and a in it too. The question only says “as a function of p” because
p is the only variable in the answer which is NOT constant.
Section 4.6: Change of Base, and Trickier Exponential Equations
There is one last algebraic law of logarithms we’ll see. It’s especially useful for calculators, when you want
to work out a log that is not base 10 or base e.
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. For instance,
log3 (12) =
ln(12)
log(12)
=
(the first uses a = e, the second uses a = 10)
ln(3)
log(3)
Ex 3: Use the Change of Base formula to simplify
log(16)
log(2) .
This law raises the question: what do we do with exponential equations where we CAN’T make one common
base? The answer is: we may use any logarithm we’d like on both sides. We use our log laws (ESPECIALLY
the Power Law) to simplify and solve for x.
Note that you have to use the same logarithm on each side!
Ex 4: Use natural logarithms to solve the equation 33x+12 = 73−9x .
NOTE: In this case, 7 is not a power of 3, so we can’t use the methods of Sections 4.2 and 4.3. Instead, the
first step applies ln to both sides.
Exponential Quadratics (aka “Quadratics in Disguise”)
Consider an equation like
e2x − 3ex + 2 = 0 i.e. (ex )2 − 3(ex ) + 2 = 0
This is a mixture of quadratic and exponential: more precisely, an exponential is plugged into a quadratic
equation! (The coefficients are a = 1, b = −3, c = 2.) We say the left side is quadratic in ex . (The usual
expression ax2 + bx + c is merely said to be quadratic in x.)
Other examples:
• 5 · 22x − 6 · 2x + 2 = 0 is quadratic in 2x .
NOTE: Keep the base and coefficient separate! 5 · 22x is not 102x .
• 21 (ln x)2 + 3 ln x = 0 is quadratic in ln x.
We’ll see this behavior of putting a function inside an equation type often. I call these expressions “quadratics
in disguise”.
How To Solve a “Quadratic in Disguise”:
1. Pick a NEW variable to stand for the exponential, like u = ax .
2. Rewrite your equation using u; it should be a normal quadratic in u.
3. Find u’s answers.
4. GO BACK to the old variable: replace your u’s by ax again and finish solving for x.
Sample: Solve e2x − 3ex + 2 = 0. Set u = ex ; we get u2 − 3u + 2 = 0. Factor: (u − 1)(u − 2) = 0, so u = 1
or u = 2. Return to old var: ex = 1 or ex = 2. Finish solving for x by using logs: x = ln(1), ln(2) .
Ex 5: Solve the following equations for x.
(a) e2x − 9ex + 77/4 = 0
(b) ex − 28e−x = −3
NOTE: To start this one, remember negative exponents are reciprocals! So write this first as: ex −28/ex =
−3. Then multiply ex through everything to get rid of the fractions.
(c) 2x + 8 · 2−x = 6
We’ll continue with “quadratics in u” next class.
Bonus: An Odd Simplification Problem (we won’t have time in class for this)
Simplify the following completely: (e3x + e−3x )2 − (e3x − e−3x )2 .
SUGGESTED STEPS:
1. Expand the squares. Remember that (e3x )2 = e6x thanks to the “power of a 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 .
NOTE: In this, all of the following skills come into play:
•
•
•
•
Squaring a sum of two terms
Taking a power of a power
Multiplying two exponentials of the same base
Thinking of negative exponents as reciprocals