HOW TO EVALUATE LOGARITHMS WITH FRACTIONS

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To evaluate logarithmic function, we should know how to convert the logarithmic function into exponential form.

Evaluate each of the following logarithms without the use of a calculator.

Problem 1 :

log4 1/2 =

Solution :

log4 1/2 = x

4x = 1/2

(22)x = 1/2

(2)2x = 2-1

2x = -1

x = -1/2

Problem 2 :

log6 1/36 =

Solution :

log6 1/36 = x

6x = 1/36

6x = 1/62

6x = 6-2

x = -2

Problem 3 :

log2/3 9/4 =

Solution :

log2/3 9/4 = x

(2/3)x = 9/4

(2/3)x = (3/2)²

(2/3)x = (2/3)-²

x = -2

Problem 4 :

log81 1/27 =

Solution :

log81 1/27 = x

81x = 1/27

(34)x = 1/33

(3)4x = 3-3

4x = -3

x = -3/4

Problem 5 :

log4 1/8 =

Solution :

log4 1/8 = x

4x = 1/8

(22)x = 1/23

(2)2x = 2-3

2x = -3

x = -3/2

Problem 6 :

log36 1/6 =

Solution :

log36 1/6 = x

36x = 1/6

(62)x = 1/6

(6)2x = 6-1

2x = -1

x = -1/2

Problem 7 :

log1/16 32 =

Solution :

log1/16 32 = x

(1/16)x = 32

(1/24)x = 25

(2)-4x = 25

-4x = 5

x = -5/4

Problem 8 :

log27 1/3 =

Solution :

log27 1/3 = x

27x = 1/3

(33)x = 1/3

(3)3x = 3-1

3x = -1

x = -1/3

Problem 9 :

log1/4 16 =

Solution :

log1/4 16 = x

(1/4)x = 16

(1/22)x = 24

(2)-2x = 24

-2x = 4

x = -4/2

x = -2

Problem 10 :

Expand the logarithmic expression :

ln (5/12x)

Solution :

= ln (5/12x)

= ln 5 - ln 12x

= ln 5 - [ln 12 + ln x]

= ln 5 - ln 12 - ln x

Problem 11 :

Use the properties of logarithms to condense each expression :

a) 3 log 4 − 2 log 𝑘

b) −5 log(𝑥 +1) + 3 log(6𝑥)

c) (1/3) log4 10 + (1/3) log4 ℎ − 6 log𝑔

d) ln (3𝑚 +5) − 4 ln 𝑚 − ln (𝑚 − 1)

e) log 20 + 2 log (1/2) − log 𝑥 + 3 log 𝑦

Solution :

a) 3 log 4 − 2 log 𝑘

= log 43  − log k2

= log 64  − log k2

= log (64/k2)

b) −5 log(𝑥 +1) + 3 log(6𝑥)

= 3 log(6𝑥) − 5 log(𝑥 + 1)

= log(6𝑥)3 − log(𝑥 + 1)5

= log(6)3 (𝑥)3 − log(𝑥 + 1)5

= log(216𝑥3) − log(𝑥 + 1)5

= log2 [(216𝑥3) / (𝑥 + 1)5]

c) (1/3) log4 10 + (1/3) log4 ℎ − 6 log𝑔

= log4 10 + log4 ℎ − log𝑔6 

= log4 (10 ℎ) − log𝑔6 

= log4 (10ℎ) − log𝑔6 

= log4 [∛(10ℎ) / 𝑔6]

d) ln (3𝑚 + 5) − 4 ln 𝑚 − ln (𝑚 − 1)

= ln (3m + 5) − ln 𝑚4 − ln (𝑚 − 1)

= ln (3m + 5) − [ln 𝑚4 + ln (𝑚 − 1)]

= ln (3m + 5) − [ln 𝑚4 (𝑚 − 1)]

= ln [(3m + 5) / 𝑚4 (𝑚 − 1)]

e) log 20 + 2 log (1/2) − log 𝑥 + 3 log 𝑦

log 20 + log (1/2)2 − log 𝑥 + log 𝑦3

log 20 + log (1/4) − log 𝑥 + log 𝑦3

log 20 ⋅ (1/4) ⋅ 𝑦3 − log 𝑥

log 5 𝑦3 − log 𝑥

log (5 𝑦3/𝑥)

Problem 12 :

Use properties of logarithms to expand each expression. The expanded logarithm expressions should have arguments with no exponent, product, or quotient

a) ln (4/5) 

b) log63x

c) log 7b/√c

d) log2(m3/8n)

e) log3[(u - 1)/(v5w3)]

Solution :

a) ln (4/5) = ln 4 - ln 5

b) log63x = log63 + log6x

c) log 7b/√c = log 7b - log √c

= log 7 + log b - log c1/2

= log 7 + log b - (1/2) log c

d) log2(m3/8n) = log2m3-  log28n

log2m3-  [log28 + logn]

= 3log2m - [log223+ logn]

= 3log2m - [3log22+ logn]

= 3log2m - [3(1) + logn]

= 3log2m - 3 - logn

e) log3[(u - 1)/(v5w3)]

= log3(u - 1) - log3(v5w3)

= log3(u - 1) - [log3v5 log3w3]

= log3(u - 1) - [5 log3v + 3 log3w]

= log3(u - 1) - 5 log3v - 3 log3w

Problem 13 :

Solve each equation.

log4 x = log4 3 + log4 (x − 2)

Solution :

log4 x - log4 (x − 2) = log4 3

log4 (x/(x − 2)) = log4 3

x/(x - 2) = 3

x = 3(x - 2)

x = 3x - 6

x - 3x = -6

-2x = -6

x = 6/2

x = 3

So, the value of x is 3.

Problem 14 :

ln(𝑥 − 3) + ln(2𝑥 + 3) = ln (−4𝑥2)

Solution :

ln(𝑥 − 3) + ln(2𝑥 + 3) = ln (−4𝑥2)

ln(x - 3)(2x + 3) = ln (−4𝑥2)

(x - 3)(2x + 3) = −4𝑥2

2x2 + 3x - 6x - 9 = −4𝑥2

2x2 + 4𝑥2 - 3x - 9 = 0

6x- 3x - 9 = 0

2x- x - 3 = 0

a = 2, b = -1 and c = -3

x = [-b ± √(b2- 4ac)]/2a

x = [1 ± √(12- 4(2)(-3))]/2(2)

= [1±√(1 + 24)]/4

= [1±√25]/4

= [1 ± 5]/4

x = (1 + 5)/4 and x = (1 - 5)/4

x = 6/4 and x = -4/4

x = 3/2 and x = -1

Problem 15 :

log (3𝑥 + 2) = 1 + 𝑙𝑜𝑔 2𝑥

Solution :

log (3𝑥 + 2) = 1 + 𝑙𝑜𝑔 2𝑥

log (3𝑥 + 2) - 𝑙𝑜𝑔 2𝑥 = 1

log [(3x + 2)/2x] = 1

(3x + 2)/2x = 101

(3x + 2)/2x = 10

3x + 2 = 10(2x)

3x + 2 = 20x

3x - 20x = -2

-17x = -2

x = 2/17

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