Soln

[a^2(3/4)b^-3(3/4)c^(3/4)]divided by (a^-1b^4c^5)

===>(a^3/2b^-9/4c^3/4)divided by (a^-1b^4c^5).

Note a^p =3/2-(-1)==>p=5/2

q=-9/4-(4)===>q=-25/4

r=3/4-(5)===>r=-17/4

now p+2q

5/2+2(-25/4)

:- p+2q=-10.

Make use of indice...

lol

How was the (a^-1b^4c^5) gotten

A comprehensible solution for you all (for those that were are finding it difficult to comprehend)

This can be solved by the application of indices..

First let me retype your question;

{(a^2.b^-3.c)^3/4}/(a^-1.b^4.c^5) = a^p.b^q.c^r

Find p + 2q.

Soln..

Please note . signifies multiplication in my solution..

So,

Multiply the numerator through with its power,

{(a^2(3/4).b^-3(3/4).c^(3/4)}/(a^-1.b^4.c^5) = a^p.b^q.c^r

Simplifying,

(a^3/2.b^-9/4.c^3/4)/(a^-1.b^4.c^5) = a^p.b^q.c^r

Further simplifying (using indices law),

a^{3/2 - (-1)}.b^(-9/4 - 4).c^(3/4 - 5) = a^p.b^q.c^r

Still simplifying,

a^{3/2 - (-1)}.b^(-9/4 - 4).c^(3/4 - 5) = a^p.b^q.c^r

a^(3/2 + 1).b^(-9/4 - 4/1).c^(3/4 - 5/1) = a^p.b^q.c^r

a^(3/2 + 1/1).b^(-9/4 - 4/1).c^(3/4 - 5/1) = a^p.b^q.c^r

a^(3 + 2)/2.b^(-9 - 16)/4.c^(3 - 20)/4 = a^p.b^q.c^r

a^5/2.b^-25/4.c^-17/4 = a^p.b^q.c^r

Now, by comparing both sides,

a^5/2 = a^p

Therefore, p = 5/2

Similarly,

b^-25/4 = b^q

Therefore, q = -25/4

Also,

c^-17/4 = c^r

Therefore, r = 17/4

But, we are looking for, p + 2q.

Hence,

p + 2q = 5/2 + 2(-25/4)

Simiplifying,

p + 2q = 5/2 + (-25/2)

Further simplifying,

p + 2q = 5/2 - 25/2

p + 2q = (5 - 25)/2

p + 2q = -20/2

p + 2q = -10

Solve fully

Hi