as maths

[raw chasm]

explain x

[maiden ermine]

Let y = 2^x

2^2x × 2^1 = 2^2x+1

Then

2y^2 - 5y -12 =0
2y^2 - 8y + 3y - 12 = 0
2y(y-4) + 3(y-4) = 0

(2y+3)(y-4) = 0

y = 4
y = -3/2

Then u can work out x

using logs

log_2(4) = x
log_2(-3/2) = x

[raw chasm]

ty! I didn’t realise how much u have to know indicies for a level

[maiden ermine]

its really necessary

youll use indices a lot

in calculus and other stuff

so make sure you have a good foundation

[maiden ermine]

oh btw theres only 1 answer for x, cuz log_2(-3/2) is complex

[raw chasm]

im kinda confused on 9b

[maiden ermine]

oh i think it says hence because y is (root3(x))^2

so use ur solutions of x from 9a

[raw chasm]

I’m confused

[maiden ermine]

what did u get for 9a

[raw chasm]

8 and

1

Are x

[maiden ermine]

@raw chasm

[raw chasm]

Ah ok

[jagged sandal]

Use a substitution

let u = 2^x

[cosmic mauve]

what this topic called