1.4 Exponents & Logarithms

Solve the equation `log_27 x = 1 - log_27 (x - 0.4)`

`log_27 (x(x - 0.4)) = 1`

`x^2 - 0.4x = 27`

`x = 5.4 or x = - 5`

`x = 5.4`

Solve the equation `log_9 81 + log_9 (1/9) + log_9 3 = log_9 x`

`log_9 81 + log_9 (1/9) + log_9 3 = log_9 [81 (1/9) 3] = log_9 27 => x = 27`

Given that `log_5 x = y,`, express each of the following in terms of y.

(a) `log_5 x^2`

(b) `log_5 (1/x)`

(c) `log_25 x`

(a)

`log_5 x^2 = 2log_5 x = 2y`

(b)

`log_5 (1/x) = - log_5 x = -y`

(c)

`log_25 x = (log_5 x) / (log_5 25) = 1/2 y`

$1000 is invested at 15% per annum interest, compounded monthly. Calculate the minimum number of months required for the value of the investment to exceed $ 3000.

15% per annum = `15/12 % = 1.25% " per month"`

Total value of investment after  months, `1000 (1.0125)^n > 3000`

`=> (1.0125)^n > 3`

`n log (1.0125) > log(3)`

`=> n > (log(3)) / (log(1.0125))`

 n > 88.4

Whole number of months required so n = 89 months.

Find the exact solution of the equation `9^(2x) = 27^(1 - x)`

`(2x) / (1 - x) = (log27) / (log9) (= 3/2)`

`4x = 3 - 3x`

`7x = 3`

`=> x = 3/7`

(a) `Let log_c 3 = p and log_c 5 = q.`Find an expression in terms of  and  for

(i) `log_c 15`

(ii) `log_c 25`

(b) Find the value of d if `log_d 6 = 1/2`

(a)

(i)

`log_c 15 = log_c 3 + log_c 5 = p + q`

(ii)

`log_c 25 = 2log_c 5 = 2q`

(b)

`d^(1/2) = 6; d = 36`

`Let p = log_10 x, q = log_10 y and r = log_10 z.`

Write the expression `log_10 (x / (y^2 sqrt(z)))` in terms of p, q and r.

`x = 10^p, y^2 = 10^(2q), sqrt(z) = 10^(r/2)`

`log_10 (x / (y^2 sqrt(z))) = log_10 (10^p / (10^(2q) 10^(r/2)))`

`= log_10 (10^(p - 2q - r/2)) (= p - 2q - r/2)`

Find the exact value of x in each of the following equations.

(a) `5^(x+1) = 625`

(b) `log_a (3x + 5) = 2`

(a) 

`5^(x+1) = 5^4`

`x + 1 = 4`

`x = 3`

(b)

Change base to give

`log(3x + 5) = log a^2`

`3x + 5 = a^2`

`x = (a^2 - 5) / 3`

(a) Find `log_2 32`

(b) Given that `log_2 ((32^x) / (8^y))` can be written as px + qy , find the value of p and of q.

(a) 5

(b)

`log_2 ((32^x) / (8^y)) = log_2 32^x - log_2 8^y`

`= x log_2 32 - y log_2 8`

`log_2 8 = 3`

p = 5; q = -3 

(accept 5x - 3y) 

Solve `log_2 x + log_2 (x - 2) = 3, "for" x > 2`

`log_2 (x(x - 2)), x^2 - 2x`

recognizing `log_a b = x <=> a^x = b` (seen anywhere)

2³ = 8

correct simplification

`x(x - 2) = 2^3, x^2 - 2x - 8`

evidence of correct approach to solve

e.g. factorizing, quadratic formula

correct working

`(x - 4)(x + 2), (2 +- sqrt(36)) / 2 x = 4`