Question

Please helphow do you solve part e??

Answer 1

q = -1 ; q= 3

There are two ways, either you can use Log or exponent rule (a^m)^n = a^(m × n):

I'll be using the exponent rule.

4^(q²-3) - 16^q = 0

It can also be written as:

4^(q²-3) - (4^2)^q = 0

Thus: 4^(q²-3) - 4^2q = 0

4^(q²-3) = 4^2q

since the base is same, comparing the exponents

(q²-3) = 2q

q² - 2q - 3 = 0

Using mid term factorisation

q² - 3q + q - 3 = 0

q (1q - 3) + 1 (q - 3) = 0

(q-1) (q-3) = 0

Thus, q = - 1

q= 3

Answer 2

q = 3 , q = - 1

Explanation:

using the rule of logarithms

log
`x^(n)` = nlogx

`4^(q^2-3)` -
`16^(q)` = 0 , then

`4^(q^2-3)` -
`4^(2q)` = 0

(q² - 3)log4 - 2qlog4 = 0 ← factor out log4 from each term

log4(q² - 3 - 2q) = 0

equate each factor to zero and solve for q

log4 ≠ 0

q² - 2q-3 = 0

(q - 3)(q + 1) = 0 ← in factored form

equate each factor to zero and solve for q

q - 3 = 0 ⇒ q = 3

q + 1 = 0 ⇒ q = - 1