indices

[quasi sierra]

do you know how to do logs?

[dusty idol]

you can split up 9 as 8+1 which is also 2^3+1, and you can rewrite 32 as 2^5

[quasi sierra]

oh true ignore me

you can also use logs

[plucky birch]

yh the solution i saw had logs

which i dk why it did cus its fm

[plucky birch]

ohh alright

[dusty idol]

2^m(1+2^n-m)=8+1/2^5

it's quite hard to visualise on text, give me a moment and i'll write it out

no logs needed I don't think

[plucky birch]

thank you

[dusty idol]

Going from line 3 to 4, I’ve rewritten 1/2^5 as 2^-5

Then if you look at both of the sides, it’s easier to compare once they’re both in the same form

[quasi sierra]

i would have never thought of doing it like that

[plucky birch]

where would you get 2^m(1+2^n-m) from?

[dusty idol]

I’ve factorised out 2^m, so if you expand out the brackets you get 2^m(1+2^n-m)—>2^m+2^n

Since multiplying out the indices will result in adding the powers so n-m+m=n

[quasi sierra]

clever how did you know how to do it like that may i ask?

like you just saw it and knew to factorise it like that

[dusty idol]

Lmao I thought logs wouldn’t be involved since the OP appears to be in y10 but the factorisation part was mainly judging different cases

[quasi sierra]

that’s so hard for year 10

[dusty idol]

And 32 being there was very convenient for the question writers, along with the fact that 9 is one more than a power of 2

[quasi sierra]

yeah true

[dusty idol]

Yeah ngl

[plucky birch]

ah i see, thank you

[dusty idol]

Nws

[mortal pecan]

Couldn’t you also rewrite 9/32 as (2³ + 2⁰)/2⁵ and then multiply both sides by 2⁵ and expand and compare

[dusty idol]

Yeah I suppose you could, how would you know which way round the values were in that case?

It’d still work for this one since it’s asking for mn though so yeah

[mortal pecan]

The values for m and n are interchangeable since it’s only one equation

[dusty idol]

Yh that’s true

They can’t really ask for the individual values here as you said