Question 1 :
Let log a = p and log b = q, write the following expressions in terms of p and q.
(i) log(ab^{3})
(ii) log(a/√b)
Answer :
Part (i) :
Using product rule of logarithm,
log(ab^{3}) = log a + log(b^{3})
Using power rule of logarithm,
= log a + 3log b
Substitute log a = p and log b = q.
= p + 3q
Part (ii) :
Using quotient rule of logarithm,
log(a/√b) = log a - log(√b)
= log a - log(b^{1/2})
Using power rule of logarithm,
= log a - (1/2)log b
Substitute log a = p and log b = q.
= p - (1/2)q
= p - q/2
Question 2 :
Let log_{4}6 = p and log_{4}11 = q. Find an expression in terms of p and q for log_{11}6.
Answer :
log_{11}6 = log_{4}6 ⋅ log_{11}4
= (log_{4}6)/(log_{4}11)
= p/q
Question 3 :
Find the value of d if log_{d}5 = 1/3.
Answer :
log_{d}5 = 1/3
The above equation is in logarithmic form. Convert it to exponential form to solve for d.
5 = d^{1/3}
Raise the exponent to 3 on both sides.
(d^{1/3})^{3} = 5^{3}
Using power of a power rule,
d^{1} = 125
d = 125
Question 4 :
Find the value of x :
log_{2a}(2x + 4) = -1
Answer :
log_{2a}(2x + 4) = -1
The above equation is in logarithmic form. Convert it to exponential form to solve for d.
2x + 4 = 5 = (2a)^{-1}
2x + 4 = 5 = 1/(2a)
Subtract 4 from both sides.
2x = 1/(2a) - 4
2x = (1 - 8a)/(2a)
Divide both sides by 2.
x = (1 - 8a)/(4a)
Question 5 :
Find the value of x :
log_{8}x = log_{8}16 + log_{8}(1/4) - log_{8}128
Answer :
log_{8}x = log_{8}16 + log_{8}(1/4) - log_{8}128
log_{8}x = log_{8}(16 ⋅ 1/4) - log_{8}128
log_{8}x = log_{8}4 - log_{8}128
log_{8}x = log_{8}(4/128)
log_{8}x = log_{8}(1/32)
Two logarithms are equal with the same. So, the arguments can be equated.
x = 1/32
Question 6 :
Find the value of x :
log_{16}36 = 1 - log_{4}x
Answer :
log_{16}36 = 1 - log_{4}x
log_{4}36 ⋅ log_{16}4 = log_{4}4 - log_{4}x
(log_{4}36)/(log_{4}16) = log_{4}(4/x)
(log_{4}6^{2})/(log_{4}4^{2}) = log_{4}(4/x)
(2log_{4}6)/(2log_{4}4) = log_{4}(4/x)
(2log_{4}6)/(2 ⋅ 1) = log_{4}(4/x)
(2log_{4}6)/2 = log_{4}(4/x)
log_{4}6 = log_{4}(4/x)
Two logarithms are equal with the same. So, the arguments can be equated.
6 = 4/x
Multiply both sides by x.
6x = 4
Divide both sides by 6.
x = 2/3
Question 7 :
It is thought that a computer virus would spread in an office according to an exponential model. 6 computers were infected initially. The number of computers infected hearing it grows at a rate of 11% per hour.
(a) How many computers were infected after 24 hours?
There are 8,200 computers in the office.
(b) How long it would take for all computers in the office to be infected.
Answer :
Part (a) :
From the given information, we can write the function which gives the number of computer infected after t hours.
I(t) = 6(1 + 0.11)^{t}
I(t) = 6(1.11)^{t}
Substitute t = 24.
I(24) = 6(1.11)^{24}
I(24) ≈ 73
After 24 hours, about 73 computers were infected.
Part (b) :
I(t) = 6(1.11)^{t}
6(1.11)^{t }= I(t)
Substitute I(t) = 8200.
6(1.11)^{t }= 8200
Divide both sides by 6.
1.11^{t }= 4100/3
Take logarithm on both sides.
log(1.11^{t}) = log(4100/3)
tlog1.11 = log4100 - log3
t(0.0453) = 3.6128 - 0.4771
0.0453t = 3.1357
Divide both sides by 0.0453.
t ≈ 69
It would take about 69 hours for all computers in the office to be infected.
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