9.5 Exponential Equations & Inequalities Logarithmic vocabulary Consider: log 260 = log (2.6 × 102) = log 2.6 + log 102 = 0.4150 +
Download ReportTranscript 9.5 Exponential Equations & Inequalities Logarithmic vocabulary Consider: log 260 = log (2.6 × 102) = log 2.6 + log 102 = 0.4150 +
9.5 Exponential Equations & Inequalities
Logarithmic vocabulary Consider: log 260 = log (2.6 × 10 2 ) = log 2.6 + log 10 2 = 0.4150 + 2 = 2.4150
Also: log 0.26
= log (2.6 × 10 –1 ) = log 2.6 + log 10 –1 = 0.4150 + –1 = –0.5850
mantissa Ex 1) Underline the mantissa & circle the characteristic log 425 = 2.6284
characteristic If we are given log
x
We will use 10
x
or
e x
.
or ln
x
, we can find
x
using our calculators. Let’s practice some simple ones…get those calculators ready!
Ex 2) Solve for
x
to the nearest hundredth.
a) log
x
= 3.2274
b) 2 log
x
= 2.6419
10 3.2274
=
x
log
x
= 1.32095 1688.11 =
x
10 1.32095
=
x
20.94 =
x
c) ln
x
– 3 = 5.7213
ln
x
= 8.7213
e
8.7213
=
x
6132.15 =
x
To solve an exponential equation: (1) Isolate the exponential expression (2) Take the logarithm of both sides of the equation (3) Verify all answers! (by substitution in original)
Ex 3) Solve for
x
to nearest hundredth.
4 2
x
– 1 – 27 = 0 4 2
x
– 1 = 27 log 4 2
x
– 1 = log 27 (2
x
– 1) log 4 = log 27 2
x
log 4 – log 4 = log 27 2
x
log 4 = log 27 + log 4
x
= log 27 + log 4 2 log 4
x
= 1.69
Ex 4)
e x e x
1 5
e x
5
e x
5 5 4
e x
5 4
e x
ln 5 4 ln 5 4 0.22
ln
e x x x
Sometimes we can’t solve algebraically, so we go to our graphing calculator.
Solve using a graphing calculator.
Ex 5)
e x
=
x
2 – 1
Y
1 =
e x Y
2 =
x
2 – 1 (Find intersection)
x
= –1.15
Ex 6)
y
≥
e x
– 2
Y
1 =
e x
– 2
Applications Compound Interest Formula:
A
P
1
r n nt A
= total value of investment
t
= number of years
P
= principal amount invested
r
= interest rate (% decimal)
n
= number of times per year interest is compounded
Ex 7) The Smith Family wants to give their youngest daughter $20,000 when she is ready for college. They now have $11,500 to invest. Determine how many years it will take them to achieve their goal given that they invest this amount at 8.3% compounded monthly.
A
P
1
r n
nt A
= 20,000
P
= 11,500
r
= .083
n
= 12 log 20, 000 11,500 log .083
12
t
12 1 .083
12
t
12 .083
12 .083
12
t
*Watch those parentheses!
t
= 7 years
Continuous Compound Interest Formula
A
=
Pe
rt
Ex 8) A sum of money invested at a fixed interest rate, compounded continuously, tripled in 19 years. Determine the interest rate at which the money was invested.
*you don’t know
A
or
P
but you don’t need it!
You need P to triple
3
A
=
Pe
rt
P
=
Pe
r
(19)
P P
3 =
e
19
r