The Student Room Group
"Given that 2^x = 1/root2 and 2^y = 4root2

(a) find the exact value of x and the exact value of y,

(b) calculate the exact value of 2^y-x"

a) 1/root2 = 1/(2^[1/2]) = 2^[-1/2] = 2^x --> x = -1/2

4root2 = 2^2 times 2^[1/2] = 2^[5/2] = 2^y --> y = 5/2

b) 2^[y-x] = 2^[5/2 - (- 1/2)] = 2^[6/2] = 2^3 = 2 times 2 times 2 = 8

I hope that's right, and helpful as well. Maths by keyboard seems a lot harder than maths in any other way.
Reply 2
Fenchurch
Given that 2^x = 1/root2 and 2^y = 4root2

(a) find the exact value of x and the exact value of y,

(b) calculate the exact value of 2^y-x

Rep given!

(a)
1/root2 = 2^-0.5 so x=-0.5
4root2 = 2^2 x 2^0.5 = 2^2.5 so y=2.5

(b)
2^y-x = 2^(2.5--0.5) = 2^3 = 8
Reply 3
hey there,

(a) we know 2^x = 1/root 2.

taking logs of both sides gives

ln 2^x = ln [1/root2]

=> x ln 2 = ln 1 - ln [root 2]

= 0 - ln [root 2]

hence, x = -ln [ root 2 ] / ln 2 .

You may wish to chnage this a little more, but that answer is an exact value of x.

(b) first, lets calculate y .

Again taking logs,

ln 2^y = ln (4root 2)

=> y ln 2 = ln (4 root 2)

=> y = ln (4root2) / ln 2

Now lets consider y - x .

y - x = ln (4 root 2)/ln 2 - ( - ln (root2))/ ln 2

= [ln (4root2) + ln (root2)] / ln 2

= [ ln(4root2 * root2)]/ ln 2

= ln 8 / ln 2

so 2^ y-x = 2^ (ln 8/ ln 2)

I think this is correct. If there is an error let me know. All I've done is just use the laws of logs for the steps anyway !!!

Hope this helps
Reply 4
i think logs is beyond C1 - I certainly haven't done it yet :P
Reply 5
R.J.A
i think logs is beyond C1 - I certainly haven't done it yet :P


yup, logs is C2, for edexcel at least.
Reply 6
I didnt realise this was C1 (sorry I should have read the title).

Apologies :s-smilie:
Reply 7
beast
I didnt realise this was C1 (sorry I should have read the title).

Apologies :s-smilie:


this way they get to see the amazing laws of logs early :wink:

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