Question

express x in terms of S...

Answer 1

where's the expression?

Answer 2

\[S= (1/2)(2^{x}+2^{-x})\] there @lgbasallote

Answer 3

so what do you think will be your first step

Answer 4

make it 2S but i cant go any further than that D:

Answer 5

so
\[2S = 2^x + 2^{-x}\]

Answer 6

you need to get x below...so you take the ln
\[\ln (2S) = \ln (2^x + 2^{-x})\]
are you familiar with ln?

Answer 7

what about log?

Answer 8

well you can use log..makes no difference really
\[\log (2S) = \log(2^x + 2^{-x})\]
got that?

Answer 9

yep...

Answer 10

so now use log laws...which do you think is applicable?

Answer 11

okay im getting confused now too sorry

Answer 12

same...

Answer 13

i ended up getting like log2^0

Answer 14

i ended up finally getting x=log2S

Answer 15

that'sonly if 2^x = S

Answer 16

yeh... i can give you the answer if you want... i just need the working out...

Answer 17

Mind if I jump in here? I have an idea.

Answer 18

by all means

Answer 19

First, look at \[2S=2^{x}+2^{-x}\]Multiply by \(2^x\). We get \[2S2^{x}=2^{2x}+1\]This is basically a quadratic equation. \[2^{2x}-2S\cdot2^x+1=0\]

Answer 20

hmmm

Answer 21

the final answer is ... \[x=\log_{2}( S+ \sqrt{S^{2}-1})\]

Answer 22

Where the polynomial is in \(2^x\). However, after this, I think we need the quadratic formula.

Answer 23

hmm...that doesnt sound simple

Answer 24

yes and @KingGeorge it did mention something about a quadratic...

Answer 25

Use the quad. formula, we get \[\Large 2^x=\frac{2S \pm \sqrt{4S^2-4}}{2}=\frac{2S\pm2\sqrt{S^2-1}}{2}=S\pm\sqrt{S^2-1}\]

Answer 26

this is why he's a genius...

Answer 27

Since \(2^x\) is never negative, we can throw out the negative, take log base 2, and there's the solution.

Answer 28

haha i am surely confused....

Answer 29

Actually, we can't throw out the negative. So there are actually two solutions. One solution would be \[x=\log_2(S+\sqrt{S^2-1})\]And the other is\[x=\log_2(S-\sqrt{S^2-1})\]

Answer 30

That can't be. There must be a reason we discard the negative solution.

Answer 31

it says in the case where x>1

Answer 32

Oh. In that case we can always throw out the negative solution because \(S-\sqrt{S^2-1}\) is in between 0 and 1, so \[\log_2(S-\sqrt{S^2-1})<1\]

Answer 33

My understanding is that the second solution will always exist under the condition that \[\sqrt{S^2-1} < S\]

Answer 34

That's correct, but \(0<S-\sqrt{S^2-1}\leq1\), and since we're told that \(x>1\), we can throw that one out.

Answer 35

Oh I see now, I missed that part

Answer 36

how do we find out xbecause isn't the quadratic formula relating to the 2?

Answer 37

After using the quadratic formula we have the relation\[\Large 2^x=S-\sqrt{S^2-1}\]So we just take log base 2 of both sides, and we get \[\Large x=\log_2(S-\sqrt{S^2-1})\]

Answer 38

OOOOHHHH i get it fuly now... thankyou

Answer 39

You're welcome.