Pretty Simple Indices equation that's eluding me

MmmPancakes

Hey guys,

So I got this homework, our teacher didn't really explain this in class yet. I think she just gave it to us for the sake of giving it to us, or maybe she didn't realise we hadn't covered it in enough detail yet.

I need your help with this


I've tried like 4 times and haven't arrived to anything conclusive, from what I can gather you let 2^x equal to y and turn it into a quadratic, and obviously you have to do this with the 2^2-x also, but it's this part that's turning me off the question. I have no idea what that value would be in y.

Any help would be appreciated, the answer is 2. (I'd post my attempts but my phone is dead and they don't lead to anything at all, I'm genuinely stuck.)

Any help figuring out this would let me do the next 8 or so she gave us by ourselves.

Cheers

EoghanIRL

MmmPancakes wrote: »

Hey guys,

So I got this homework, our teacher didn't really explain this in class yet. I think she just gave it to us for the sake of giving it to us, or maybe she didn't realise we hadn't covered it in enough detail yet.

I need your help with this


I've tried like 4 times and haven't arrived to anything conclusive, from what I can gather you let 2^x equal to y and turn it into a quadratic, and obviously you have to do this with the 2^2-x also, but it's this part that's turning me off the question. I have no idea what that value would be in y.

Any help would be appreciated, the answer is 2. (I'd post my attempts but my phone is dead and they don't lead to anything at all, I'm genuinely stuck.)

Any help figuring out this would let me do the next 8 or so she gave us by ourselves.

Cheers

Look in your log tables . There is a section for the rules of logs and indices.
Use logs when x is in the power.

ZorbaTehZ

Since [latex]2^{2-x}=2^2\cdot 2^{-x}=\frac 4 {2^x}[/latex] if you let [latex]y=2^x[/latex] then the quadratic you get is [latex]y^2-5y+4[/latex]; solve that and then substitute back in for [latex]x[/latex].

MmmPancakes

ZorbaTehZ wrote: »

Since [latex]2^{2-x}=2^2\cdot 2^{-x}=\frac 4 {2^x}[/latex] if you let [latex]y=2^x[/latex] then the quadratic you get is [latex]y^2-5y+4[/latex]; solve that and then substitute back in for [latex]x[/latex].

oh wow, I was totally overcomplicating it

Thanks a bunch