Daily study session 2:

Hi all students, the questions for the daily study session 2 is available now.

Please try out and send me the answer

 

Answer:

Q1

(a)

 2^x=y

2^(x+3) = (2^x) (2^3) = y(2^3) = 8y

(b)

4[(2^3)^(1-x)] = 4[8^(1-x)]

=4[(8^1)/(8^x)]

=4[8((2^3)^x)]

=4[8((2^x)^3)]

=4[8(y^3)]

=32(y^3)

(c)

8[4^(x-2)]

=8[(4^x)/(4^2)]

=8[((2^2)^x)/16]

=8[((2^x)^2)/16]

=8[(y^2)/16]

=(y^2)/2

(d)

(8^x) - (4^-x)

= [(2^3)^x]-[(2^2)^(-x)]

=[(2^x)^3]-[(2^x)^(-2)]

=(y^3)-(y^(-2))

= (y^3)-[1/(y^2)]

Q2

{[8(x^2)]^(8-r)} {[1/(2x)]^r}

={[(2^3)(x^2)]^(8-r)}{[(2x)^(-1)]^r}

={(2^3)^(8-r)}{(x^2)^(8-r)}{2^(-r)}{x^(-r)}

=[2^(24-3r)][ x^(16-2r)][2^(-r)][x^(-r)]

=[2^(24-3r-r)][x^(16-2r-r)]

=[2^(24-4r)][x^(16-3r)] (shown)