Logs & Exps Solutions

1. a) x = log₂2 b) log₂x = 5 c) -2 = log₃x

2^x = 2 2⁵ = x 3^{-2 =} x

x = 1 x = 32 x = 1/9

2. a) log₃81 b) log₄1 c) log₅()

let x = log₃81 let x = log₄1 let x = log₅()

3^x = 81 4^x = 1 5^x = 1/125

x = 4 x = 0 x = -3

3. a) log₁₀5 + log₁₀10 – log₁₀(1/2)

log₁₀(5x10) – log₁₀(1/2)

log₁₀50 - log₁₀(1/2)

log₁₀(50/0.5)

log₁₀(100)

= 2 since 10² = 100

b) log 5 + log 6 – log10 + log(1/3)

log30 –log10 + log(1/3)

log3 + log(1/3)

log1

= 0

c) log₃9 – log₃(1/3)

log₃27

= 3

4. a) log(x + 1) + log(x – 1) = log3

log= log3

= 3

x + 1 = 3x – 3

x = 2

b) log₅(x + 1) + log₅(x – 3) = 1

log₅ = log₅5 (note that log₅5 = 1)

= 5

x + 1 = 5x – 15

x = 4

c) 2log₂x + log₂3 = 7

log₂x² + log₂2 = log₂128 (note log₂128 = 7)

log₂2x² = log₂128

2x² = 128

x² = 64

x = 8

5.

Find the relationship between x and y.

Since both the axis are logs the relationship is of the form

y = ax^b

logy = logax^b

log₁₀y = log₁₀a + log₁₀x^b

log₁₀y = log₁₀a + blog₁₀x

this is of the form:

Y = bX + c

from the graph

c = log₁₀a

0.2 = log₁₀a

a = 10^0.2

a = 1.58 b is simply the gradient of the line = 3/2

Solution: y = 1.58(x)^1.5

6.

Find the relationship between x and y.

This is of the form

y = ab^x

log₁₀y = log₁₀a + log₁₀b^x

log₁₀y = xlog₁₀b + log₁₀a

This is of the form

Y = mx + C

gradient m = 0.3 = log₁₀b

b = 10^0.3

b = 2

Y intercept: log₁₀a = 0.5

a = 10^0.5

a = 3.16

Solution

y = 3.16(2)^x

7 a) P = 80 000(1.2)^0.1t if t = 20 then 0.1t = 2

P = 80 000(1.2)²

P = 115 200

b) 160 000 = 80 000(1.2)^0.1t

2 = 1.2^0.1t

ln 2 = ln(1.2)^0.1t

ln 2 = 0.1t x ln(1.2)

0.1 t = ln(2) / ln(1.2)

t = 38 years