Section 3.4: Exponential and Logarithmic Equations
There are two approaches to take when solving equations involving exponents and
logarithms:
1. One‐to‐One properties
2. Cancelation properties
1. Using One‐to‐One properties to solve
If f (x) is a one‐to‐one function then when f (x) = f (a) it implies x = a .
For instance, if 2 x = 212 then that implies x = 12 and the equation is solved.
Rewrite the exponential equation in a form like this and you can solve it.
ex) Solve these equations:
a) 53 x+2 = 5x−7
2
b) 10 x +2 x =
1
1000
c) 274 x−1 = 812 x
The same logic applies to some logarithmic equations:
log a ( x) = log a (b) implies x = b.
ex) Solve ln(x + 13) − ln(4) = ln(3x − 5) (HINT: Condense the left side first)
2. Cancelation Properties
When the one‐to‐one properties aren’t available you’ll need to use the cancelation
properties to solve.
When the variable expression is in the exponent you’ll need to get the
exponential term completely isolated on the left side:
aX = b
... apply the appropriate logarithm to both sides
log a (a x ) = log a (b)
x
log a ( a ) = log a (b)
x = log a (b)
... the cancelation property kicks in ...
... and this ‘frees’ the variable from the exponent.
ex) Solve the following exponential equations. First give an EXACT answer then
approximate the answer to 4 decimal places.
a) 4 3 x−5 = 230
b) 6 + 2e x / 3 = 24
c)
1500
= 200
20 − 5e 0.1 x
(remember to get the exponential term isolated first)
When solving logarithmic equations you’ll first need to get a single logarithmic
expression on the left side (this may require using the laws I, II and III)
log a (x) = B
aloga ( x ) = aB
a
log a ( x )
= aB
... “exponentiate” both sides to use the cancel property
... this frees the variable from the logarithm, then solve.
x = aB
The resulting equation may be linear, quadratic, etc.
YOU MUST CHECK FOR EXTRANEOUS SOLUTIONS!!
ex) Solve the following logarithmic equations. First give an EXACT answer then
(when applicable) approximate the answer to 4 decimal places.
a) ln(2 x + 5) = 4
c) log( x) + log(x − 3) = 1
b) log2 (x + 9) − log2 (x − 5) = 3