Download Math 126: Study List Test 3 (Solutions) Exponential and Logarithmic

Math 126: Study List Test 3 (Solutions)
Exponential and Logarithmic functions
Use a calculator to evaluate each function.
1. ln(30)
Solution: ln(30) ≈ 3.40
2. e4
Solution: e4 ≈ 54.6
log(A)
3. log6 60 Hint: Use the change of base formula. logB A = log(B)
Solution: log6 60 =
log(60)
log(6)
≈ 2.285
4. e−1
Solution: e−1 ≈ .368
5. log(50000)
Solution: log(50000) ≈ 4.699
1
Graph each function
1. y = ex
2. f (x) = 2x
2
3. f (x) = ex − 1
Solve each equation
1. 32x−1 = 81
Solution:
32x−1 = 81
32x−1 = 34
2x − 1 = 4
2x = 5
2x
5
2 = 2
x = 52
1
2. 3x+2 = 243
Solution:
1
3x+2 = 243
3x+2 = 315
3x+2 = 3−5
x + 2 = −5
x = −5 − 2
x = −7
3
3. 8x = 90
Solution:
8x = 90
ln(8x ) = ln(90)
x · ln(8) = ln(90)
x = ln(90)
ln(8)
x ≈ 2.07
4. ex = 27
Solution:
ex = 27
ln(ex ) = ln(27)
x = ln(27)
x ≈ 3.30
Convert each logarithm to an exponential equation
1. log7 49 = 2
Solution:
72 = 49
2. ln(c) = d
Solution:
ed = c
4
Convert each exponential equation to a logarithm
1. 43 = 64
Solution: log4 64 = 3
2. ex = y
Solution: ln(y) = x
3. 103 = 1000
Solution: log(1000) = 3
Derivatives of logarithms and exponential functions
Find the derivative of each function
1. f (x) = ln(x) + ex + x4
Solution:
f (x) = ln(x) + ex + x4
f ′ (x) = x1 + ex + 4x3
2. g(x) = ln(4x5 )
Solution:
g(x) = ln(4x5 )
5 ′
)
g ′ (x) = (4x
4x5 =
20x4
4x5
=
5
x
5
2
3. f (x) = e4x
Solution:
2
f (x) = e4x
2
f ′ (x) = 8xe4x
4. f (x) = 4x3 ex
Solution:
f (x) = 4x3 ex
f ′ (x) = (ex )′ 4x3 + (4x3 )′ ex
f ′ (x) = 4x3 ex + 12x2 ex
5. f (x) = x5 ln(x)
Solution:
f (x) = x5 ln(x)
f ′ (x) = (ln(x))′ · x5 + (x5 )′ · ln(x)
f ′ (x) = x1 · x5 + 5x4 ln(x)
f ′ (x) = x4 + 5x4 ln(x)
3
6. y = x5 e6x
Solution:
3
y = x5 e6x
3
3
y ′ = (e6x )′ x5 + (x5 )′ e6x
3
3
y ′ = 18x2 e6x · x5 + 5x4 e6x
3
3
y ′ = 18x7 e6x + 5x4 e6x
6
2
7. y = ex
+2x+3
Solution:
2
y = ex +2x+3
2
y ′ = (2x + 2)ex +2x+3
8. f (x) = 3x2 ln(4x)
Solution:
f (x) = 3x2 ln(4x)
f ′ (x) = (ln(4x))′ 3x2 + (3x2 )′ ln(4x)
4
· 3x2 + 6xln(4x)
f ′ (x) = 4x
f ′ (x) = 3x + 6xln(4x)
9. g(x) = ln(x4 )
Solution:
g(x) = ln(x4 )
3
4
g ′ (x) = 4x
x4 = x
ln(x)
10. f (x) = x3
Solution:
x3 ·(ln(x))′ −ln(x)(x3 )′
(x3 )2
x3 · 1 −ln(x)·(3x2 )
= x (x3)2
x2 −3x2 ln(x)
=
x6
f ′ (x) =
f ′(x)
f ′(x)
7
Find the slope of the tangent line to the given function at the given
point.
1. f (x) = ex : (0, 1)
Solution:
f (x) = ex
f ′ (x) = ex
Slope: m = f ′ (0) = e0 = 1
2. f (x) = xex − 1: (0, −1)
Solution:
f (x) = xex − 1
f ′ (x) = (ex )′ x + (x)′ ex = xex + 1 · ex = xex + ex
Slope: m = f ′ (0) = 0 · e0 + e0 = 0 + 1 = 1
8