MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Fundamentals of Mathematics
(MATH 1510)
Instructor: Lili Shen
Email: [email protected]
Department of Mathematics and Statistics
York University
January 25-29, 2016
Outline
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
1
Exponential and Logarithmic Equations
Exponential equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
An exponential equation is one in which the variable occurs
in the exponent.
Some exponential equations can be solved by using the fact
that exponential functions are one-to-one, i.e.,
ax = ay ⇐⇒ x = y.
Exponential equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Example
Solve the exponential equation
3x+2 = 7.
Exponential equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Solution.
3x+2 = 7,
log3 3x+2 = log3 7,
x + 2 = log3 7,
x = log3 7 − 2.
Exponential equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
In this example,
x = log3 7 − 2
is the exact solution of the equation 3x+2 = 7.
Using calculators, one could find an approximation to the
exact solution by the change of base formula:
x = log3 7 − 2 =
ln 7
− 2 ≈ −0.228756.
ln 3
Exponential equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Example
Solve the equation
e2x − ex − 6 = 0.
Exponential equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Solution.
e2x − ex − 6 = 0,
(ex − 3)(ex + 2) = 0,
ex = 3
or
ex = −2.
Since ex > 0, one could only have ex = 3 and consequently
x = ln 3.
Logarithmic equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
A logarithmic equation is one in which a logarithm of the
variable occurs.
Some exponential equations can be solved by using the fact
that logarithmic functions are one-to-one, i.e.,
loga x = loga y ⇐⇒ x = y.
Logarithmic equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Example
Solve each equation for x.
(1) ln x = 8.
(2) log2 (25 − x) = 3.
Logarithmic equations
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Solution.
(1)
eln x = e8 ,
x = e8 .
(2)
log2 (25 − x) = 3,
2log2 (25−x) = 23 ,
25 − x = 8,
x = 17.
Compound interest
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
If a principal P is invested at an annual interest rate r for a
period of t years, then the amount A of the investment is
given by
A(t) = P(1 + rt) (simple interest),
r nt
A(t) = P 1 +
(interest compounded n times per year),
n
A(t) = Pert (interest compounded continuously).
Compound interest
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Example
A principal P is invested at an interest rate r per year. Find
the time required for the money to double if the interest is
(1) simple,
(2) compounded semiannually,
(3) compounded continuously.
Compound interest
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
Solution.
Suppose that the money will be double after t years.
(1) Simple interest:
P(1 + rt) = 2P,
1 + rt = 2,
1
t= .
r
Compound interest
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
(2) Interest compounded semiannually:
r 2t
2
r 2t
(1 +
2
r 2t
log1+ r (1 +
2
2
2t
P 1+
= 2P,
= 2,
= log1+ r 2,
2
= log1+ r 2,
2
1
t = log1+ r 2.
2
2
Compound interest
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
(3) Interest compounded continuously:
Pert = 2P,
ert = 2,
ln ert = ln 2,
rt = ln 2,
ln 2
t=
.
r
Compound interest
MATH 1510
Lili Shen
Exponential
and
Logarithmic
Equations
For example, if the annual interest rate is 5%, the number of
years t needed to double an investment is:
(1) Simple interest:
t=
1
1
=
= 20.
r
0.05
(2) Interest compounded semiannually:
t=
1
ln 2
log1+ r 2 =
≈ 14.04.
2
2
2 ln 1.025
(3) Interest compounded continuously:
t=
ln 2
ln 2
=
≈ 13.86.
r
0.05