Exponential
See how to solve an exponential equation/inequality/function/change.
13 examples and their solutions.
Exponential Equation
Example
23x - 1 = 4
Solution 23x - 1 = 4
23x - 1 = 22
3x - 1 = 2 - [1]
3x = 3
x = 1
23x - 1 = 22
3x - 1 = 2 - [1]
3x = 3
x = 1
[1]
23x - 1 = 22
The bases are the same: 2.
Then the exponents are the same.
The bases are the same: 2.
Then the exponents are the same.
Close
Example
3x - 5 = 94
Solution Example
74x + 8 = 1
Solution 74x + 8 = 1
= 70
4x + 8 = 0
4x = -8
x = -2
= 70
4x + 8 = 0
4x = -8
x = -2
Close
Example
125⋅5x = (125)x
Solution 125⋅5x = (125)x
53⋅5x = 25-x
53 + x = (52)-x
5x + 3 = 5-2x
x + 3 = -2x
3x = -3
x = -1
53⋅5x = 25-x
53 + x = (52)-x
5x + 3 = 5-2x
x + 3 = -2x
3x = -3
x = -1
Close
Example
4x = 3⋅2x + 4
Solution 4x = 3⋅2x + 4
4x - 3⋅2x - 4 = 0
(2x)2 - 3⋅2x - 4 = 0
(2x - 4)(2x + 1) = 0 - [1]
1) 2x - 4 = 0
2x = 4
= 22
x = 2
2) 2x + 1 = 0
2x = -1( x ) - [2]
x = 2
4x - 3⋅2x - 4 = 0
(2x)2 - 3⋅2x - 4 = 0
(2x - 4)(2x + 1) = 0 - [1]
1) 2x - 4 = 0
2x = 4
= 22
x = 2
2) 2x + 1 = 0
2x = -1( x ) - [2]
x = 2
[2]
2x = -1
The left side, 2x, is (+).
The right side, -1, is (-)
So there's no solution for case 2.
The left side, 2x, is (+).
The right side, -1, is (-)
So there's no solution for case 2.
Close
Exponential Inequality
Example
25x - 9 > 4x
Solution 25x - 9 > 4x
25x - 9 > 22x - [1]
5x - 9 > 2x
3x > 9
x > 3
25x - 9 > 22x - [1]
5x - 9 > 2x
3x > 9
x > 3
[1]
The base 2 is in
2 > 1.
So the order of the inequality sign
doesn't change.
> → >
2 > 1.
So the order of the inequality sign
doesn't change.
> → >
Close
Example
116⋅(18)x ≤ (14)x
Solution 116⋅(18)x ≤ (14)x
(12)4⋅(12)3x ≤ (12)2x
(12)4 + 3x ≤ (12)2x - [1]
4 + 3x ≥ 2x - [2]
x ≥ -4
(12)4⋅(12)3x ≤ (12)2x
(12)4 + 3x ≤ (12)2x - [1]
4 + 3x ≥ 2x - [2]
x ≥ -4
[2]
The base 1/2 is in
0 < 1/2 < 1.
So the order of the inequality sign
does change.
≤ → ≥
0 < 1/2 < 1.
So the order of the inequality sign
does change.
≤ → ≥
Close
Exponential Function: Graph
Graph: y = ax (a > 1)
(0, a0) = (0, 1)
2. The asymptote of the graph is the x-axis.
(= The graph follows the x-axis.)
Graph: y = ax (0 < a < 1)
(0, a0) = (0, 1)
2. The asymptote of the graph is the x-axis.
(= The graph follows the x-axis.)
Example
Graph y = 2x.
Solution Draw (0, 1).
↓
Draw the points
when x = 1, 2, 3 and -1, -2, -3.
(1, 21) = (1, 2)
(2, 22) = (2, 4)
(3, 23) = (3, 8)
(-1, 2-1) = (-1, 1/2)
(-2, 2-2) = (-2, 1/4)
(-3, 2-3) = (-3, 1/8)
when x = 1, 2, 3 and -1, -2, -3.
(1, 21) = (1, 2)
(2, 22) = (2, 4)
(3, 23) = (3, 8)
(-1, 2-1) = (-1, 1/2)
(-2, 2-2) = (-2, 1/4)
(-3, 2-3) = (-3, 1/8)
↓
Close
Example
Graph y = (13)x.
Solution ↓
Draw the points
when x = 1, 2 and -1, -2.
(1, (1/3)1) = (1, 1/3)
(2, (1/3)2) = (2, 1/9)
(-1, (1/3)-1) = (-1, 3)
(-2, (1/3)-2) = (-2, 32) = (-2, 9)
when x = 1, 2 and -1, -2.
(1, (1/3)1) = (1, 1/3)
(2, (1/3)2) = (2, 1/9)
(-1, (1/3)-1) = (-1, 3)
(-2, (1/3)-2) = (-2, 32) = (-2, 9)
↓
Close
Exponential Growth/Decay: Final Value
Formula
A = A0(1 + r)t
A: final value
A0: initial value
r: rate of change
t: time
Compound InterestA: final value
A0: initial value
r: rate of change
t: time
Exponential Growth/Decay: Time
Example
The population of a town is 10,000. If it increases at a rate of 7% per year, find the expected population 12 years later.
(Assume 1.0712 = 2.252.)
Solution (Assume 1.0712 = 2.252.)
A0 = 10000
r = 0.07 /year
t = 12 years
A = 10000⋅(1 + 0.07)12
= 10000⋅1.0712
= 10000⋅2.252
= 22520
r = 0.07 /year
t = 12 years
A = 10000⋅(1 + 0.07)12
= 10000⋅1.0712
= 10000⋅2.252
= 22520
Close
Example
A radioactive substance weighs 80g. If it decreases at a rate of 5% per minute, find the expected weight 1 hour later.
(Assume 0.9560 = 0.046.)
Solution (Assume 0.9560 = 0.046.)
A0 = 80 g
r = -0.05 /minute
t = 60 minutes
A = 80⋅(1 - 0.05)60
= 80⋅0.9560
= 80⋅0.046
= 3.68 g
r = -0.05 /minute
t = 60 minutes
A = 80⋅(1 - 0.05)60
= 80⋅0.9560
= 80⋅0.046
= 3.68 g
Close
Continuous Exponential Growth/Decay: Final Value
Formula
A = A0ert
A: final value
A0: initial value
e: constant number (= 2.71828...)
r: rate of change
t: time
Compound InterestA: final value
A0: initial value
e: constant number (= 2.71828...)
r: rate of change
t: time
Continuous Exponential Growth/Decay: Time
Example
A substance weighs 10g. If it continuously increases at a rate of 3% per second, find the expected weight 1 minute later.
(Assume e1.8 = 6.05.)
Solution (Assume e1.8 = 6.05.)
A0 = 10
r = 0.03 /second
t = 60 seconds
A = 10⋅e0.03⋅60
= 10⋅e1.8
= 10⋅6.05
= 60.5 g
r = 0.03 /second
t = 60 seconds
A = 10⋅e0.03⋅60
= 10⋅e1.8
= 10⋅6.05
= 60.5 g
Close
Example
A radioactive substance weighs 80g. If it continuously decreases at a rate of 5% per minute, find the expected weight 1 hour later.
(Assume e-3 = 0.050.)
Solution (Assume e-3 = 0.050.)
A0 = 80
r = -0.05 /minute
t = 60 minutes
A = 80⋅e-0.05⋅60
= 80⋅e-3
= 80⋅0.05
= 4.0 g
r = -0.05 /minute
t = 60 minutes
A = 80⋅e-0.05⋅60
= 80⋅e-3
= 80⋅0.05
= 4.0 g
Close