Question

(ar^b)^4=81r^24 where a and b are positive integers. Work out a and b

Answer 1

Answer:

Vales of a =3 and b = 6

Step-by-step explanation:

Given that: $(ar^b)^4 = 81r^{24}$            .....[1] where a and b are positive integers

we can write 81 and 24 as;

$81 = 3 \cdot 3 \cdot 3 \cdot 3 = 3^4$

$24 = 4 \cdot 6$

We have [1] as;

$(ar^b)^4 = 3^4r^{4 \cdot 6}$

Using power rules;

$a^nb^n = (ab)^n$

$a^n = b^n$ which implies a = b

then;

$(ar^b)^4 = (3r^6)^4$

$ar^b = 3r^6$

On comparing both sides we have;

a = 3 and

$r^b = r^6$

⇒$b = 6$

Therefore, the value of a and b are, 3 and 6