Edexcel GCSE maths, give each other questions to do.

$\left(\dfrac{2^{-3x}}{2^{-2}}\right)^2=2^{10}$

find x

$\left(\dfrac{2^{-3x}}{2^{-2}}\right)^2=2^{10}$

find x

X= -1

thank you so much!!!

-2/12?

is it -3 ?

nope nope unfortunately

loll i deleted my comment bc i realised i forgot to square the bracket

well the topic is the same except this is a much harder version so it is actually very good practice.

well I guess we won't actually know it's been made harder untill the exam, but I figured cuz we're the last year doing this specification, they will make it harder maybe idk, because the new spec is much harder and they want to stretch us as much as they can in the current spec.

Why is it x = -1 ? help pls

That's the last question of a FM paper. Why are you giving people this in this thread? To make yourself feel better and higher?

omg, it's the same topic but a little harder, it will prepare u for normal gcse maths much more , as it requires you to think a little more, so shh.

You purposely put probably one of the hardest FM questions in order to taunt people about their abilities on an online forum. Pretty pathetic lmao

I think, with all due respect, it was more a lack of understanding on your part if you didn't know what to do, as you say you know what to do for these types of questions now.

ok
so

$\left(\dfrac{2^{-3x}}{2^{-2}}\right)^2=2^{10}$

$\dfrac{2^{-3x}}{2^{-2}}=2^5$

$2^{-3x}=2^3$

$\dfrac{1}{2^{3x}}=2^3$

$\dfrac{1}{2^3}=2^{3x}$

$\dfrac{1}{2^3}=\left(2^x\right)^3$

let $2^x$ = y

$\dfrac{1}{8}=y^3$

$\dfrac{1}{2}=y$

$2^x=\dfrac{1}{2}$

x=-1

For probability & tree diagrams: http://www.mathsgenie.co.uk/resources/92_tree-diagrams.pdf

These are A* questions for Edexcel GCSE but he charges for the solutions. :/
1H http://bland.in/a_star1h_new2.pdf
2H http://bland.in/a_star2h_new2.pdf

http://saveyourexams.co.uk/ is a site I use but you'll have to register and the edexcel GCSE maths doesn't have the question papers yet, however it does have the edexcel IGCSE maths questions + solutions by topic.