# Help!! P2 logarithm question, too tough for me

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#1
Hi, This one is a real problem for me,

If xy=64 and log(basex)y + log(basey)x = 5/2

find x and y

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After changing the base I get

(lgy/lgx) + (lgx/lgy) = 5/2

After this I keep getting it wrong? any ideas?

thanks
0
16 years ago
#2
change to one base (x or y or e or 10)
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#3
(Original post by kikzen)
change to one base (x or y or e or 10)
already done that!!! lgx equals log (base 10) x

anyways here is the answer just go it

(lgy/lgx) + (lgx/lgy) = 5/2

(2)(lgy)(lgy)+(2)(lgx)(lgx)= 5(lgx)(lgy)

let lgy=t
let lgx=v

so, 2t^2 - 5tv + 2v^2 = 0

(2t - v)(t-2v)

(1) v=2t and (2) t=2v

from (1)

lgx = 2lgy

but xy = 64

lgx + lgy = lg64

therefore 2lgy + lgy = lg64

3lgy = 3lg4

so y=4

xy = 64

so x=16

from (2)

lgy = 2lgx

but xy = 64

lgx + lgy = lg64

therefore 2lgx + lgx = lg64

3lgx = 3lg4

so x=4

xy = 64

so x=16

so the answer is (16,4) or (4,16)

I think it's because (lgx)(lgx) can only be solved by letting it equal something else like t, then factorising.
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