EVALUATE BASIC LOG FUNCTIONS WITHOUT CALCULATOR

Without using a calculator, simplify.

Problem 1 :

log 8 / log 2

Solution :

Simplifying the numerator and denominators separately.

log 8 = log2³

= 3log2

= 3log2/log2

= 3(1)

= 3

Problem 2 :

log 9 / log 3

Solution :

Simplifying the numerator, we get

= log 9 / log 3

= log 3²

= 2 log 3

Applying it in the given question, we get

= 2 log 3/log3

= 2(1)

= 2

Problem 3 :

log 4 / log 8

Solution :

= log 4 / log 8

= log 2²/log 2³

= 2/3

Problem 4 :

log 5 / log (1/5)

Solution :

= log 5 / log (1/5)

= log 5 / log 5^-1

= log 5 / -1log 5

= -(log 5/log 5)

= -1

Problem 5 :

log (0.5) / log 2

Solution :

= log (0.5) / log 2

= log (5/10)/ log 2

log (5/10) x 2

= log5/5

= 1

Problem 6 :

log 8 / log (0.25)

Solution :

= log 8 / log (0.25)

log 8 = log 2³

= 3 log 2 ----(1)

log (0.25) = log (25/100)

= log 1/4

= log 1/2²

= log 2^-2

= -2 log 2 ----(2)

(1) / (2)

= 3 log 2/(-2 log 2)

= -3/2

Problem 7 :

log 2^b / log 8

Solution :

log_b a = log a / log b

= log 2^b / log 8

= log₈ 2^b

= b × log₈ 2

= b × log₂³ 2

= b × 1/3 × log₂2

= b/3

Problem 8 :

log 4 / log 2^a

Solution :

= log 4 / log 2^a

 log 4 = log 2²

= 2 log 2 ---(1)

log 2^a = a log 2 ---(2)

(1) / (2)

= 2 log 2/a log 2

= 2/a

Problem 9 :

Under certain conditions, the wind speed s (in knots) at an altitude of h meters above a grassy plain can be modeled by the function

s(h) = 2 ln 100h

a. By what amount does the wind speed increase when the altitude doubles?

b. Show that the given function can be written in terms of common logarithms as

s(h) = (2 /log e)  (log h + 2)

Solution :

a)

s(h) = 2 ln 100h

When h = 2h

s(h) = 2 ln 100(2h)

= 2 ln (200 h)

= 2 ln (2 x 100 h)

= 2 [ln 2 + ln 100 + ln h]

= 2 ln 2 + 2 [ln 100 + ln h]

= ln 2² + 2 [ln 100 + ln h]

= ln 4 + 2 [ln 100 h]

= 1.38 + s(h)

So, the wind speed will increase by 1.38 knots.

b) s(h) = (2 /log e)  (log h + 2)

= 2 (log h + 2) / log e

= (2 log h + 4) / log e

= (log h² + 4) / log e

Solve each equation.

Problem 10 :

 -3 ln x = -24

Solution :

 -3 ln x = -24

ln x = -24/(-3)

ln x = 8

x = e⁸

Problem 11 :

4 - 3log (5x) = 16

Solution :

4 - 3log (5x) = 16

4 - 16 = 3log (5x)

-12 = 3log (5x)

-12/3 = log (5x)

-4 = log (5x)

10^-4 = 5x

1/10⁴ = 5x

1/10000 = 5x

x = 1/50000

Problem 12 :

log₃(x - 1) = -2

Solution :

log₃(x - 1) = -2

x - 1 = 3^-2

x - 1 = 1/3²

x - 1 = 1/9

x = (1/9) + 1

x = (1 + 9)/9

x = 10/9

Problem 13 :

log₂(x² - 4) = log₂ 21

Solution :

log₂(x² - 4) = log₂ 21

x² - 4 = 21

x² = 21 + 4

x² = 25

x = √25

x = -5 and 5

Problem 14 :

The function

s(d) = 0.159 + 0.118 log d

relates the slope, s, of a beach to the average diameter, d, in millimeters, of the sand particles on the beach. Which beach has a steeper slope: beach A, which has d = 0.0625, or beach B, which has very coarse sand with d = 1 ? Justify your decision. 

Solution :

s(d) = 0.159 + 0.118 log d

Calculate the slope for Beach A :

When d = 0.0625

s(0.0625) = 0.159 + 0.118 log (0.0625)

= 0.159 + 0.118(-1.204)

= 0.159 - 0.142

= 0.017

Calculate the slope for Beach B :

When d = 1

s(1) = 0.159 + 0.118 log (1)

= 0.159 + 0.118(0)

= 0.159 + 0

= 0.159

Slope for Beach B > Slope for Beach A

Problem 15 :

The function

S(d) = 93 log d + 65

relates the speed of the wind, S, in miles per hour, near the center of a tornado to the distance that the tornado travels, d, in miles.

a) If a tornado travels a distance of about 50 miles, estimate its wind speed near its center.

b) If a tornado has sustained winds of approximately 250 mph, estimate the distance it can travel.

Solution :

S(d) = 93 log d + 65

a) When d = 50 miles

S(50) = 93 log 50 + 65

= 93(1.698) + 65

= 157.91 + 65

= 222.91

b) S(d) = 93 log d + 65

s(d) = 250 mph and d = ?

250 = 93 log d + 65

250 - 65 = 93 log d

185 = 93 log d

185/93 = log s

1.98 = log s

10^1.98 = s

s = 95.4

Approximately 95 miles