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Freddy C
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#1
Report Thread starter 13 years ago
#1
got 2 questions:

1) if xy = 62 and logxy + logyx = (5/2), find x and y

2) prove that if a^x = b^y = (ab)^xy, then x+y = 1

Thanks for any answers
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Fermat
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#2
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#2
1)
logxy = 1/logyx = u

u + 1/u = 5/2
2u² + 2 = 5u
2u² - 5u + 2 = 0
(2u - 1)(u - 2) = 0
u = ½, u = 2
==========
logxy = ½
y = x^½
x = y²

xy = 64
y².y = 64
y^3 = 64
y = 4
x = 16
=====

taking u = 2 gives logyx = ½ gives

y = 16
x = 4
====
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Gaz031
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#3
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#3
(Original post by Freddy C)
got 2 questions:

1) if xy = 62 and logxy + logyx = (5/2), find x and y
lny/lnx + lnx/lny=(5/2), y=62/x:
ln(62/x)/lnx + lnx/ln(62/x)=(5/2)
[ln62-lnx]/lnx + lnx/[ln62-lnx]=(5/2)
Multiply through by lnx(ln62-lnx) to obtain a quadratic in lnx and hence solve for x, then y.

2) prove that if a^x = b^y = (ab)^xy, then x+y = 1

Thanks for any answers
Let p=a^x=b^y=(ab)^xy. Find x,y,xy in terms of a,b and p. Then substitute in for x+y to show it is equal to 1.
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El Stevo
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#4
Report 13 years ago
#4
xy = 62

logx/logy + log/logx = 5/2
let logx/logy = n

n + 1/n = 5/2
(n^2) + 1 = 5n/2
(n^2) - 5n/2 + 1 = 0
n = [5/2 +/- sqrt(25/4 -4)]2
n = [5/2 + sqrt(2.25)]/2
n = [5/2 +/- 1.5]/2

n = 2, 0.5

logx/logy = 2
logx = 2logy
x = (y^2)

(y^3)=62
y = 3.96
x = 15.7

(using n=0.5 would have given x and y the other way round - makes no difference)
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Fermat
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#5
Report 13 years ago
#5
2)
a^x = b^y = ab^(xy)

ln(a^x) = xlna
ln(b^y) = ylnb

ln(ab^(xy)) = xyln(ab) = xy(lna + lnb)

substituting for xlna = ln(a^x) = ln(ab^(xy))

xlna = xy(lna + lnb)
xlna(1 - y) = x.ylnb
xlna(1 - y) = x.xlna
(1 - y) = x
x + y = 1
=======
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