SOLUTION: Solve the following system for (x,y): {{{log(9,X)+ log(Y,8)=2}}} {{{log(X,9)+ log(8,y)=8/3}}} (log to the base 9 of X) + (log to the base Y of 8) = 2 (log to the base X of

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the following system for (x,y): {{{log(9,X)+ log(Y,8)=2}}} {{{log(X,9)+ log(8,y)=8/3}}} (log to the base 9 of X) + (log to the base Y of 8) = 2 (log to the base X of       Log On


   

Question 29446: Solve the following system for (x,y):
log%289%2CX%29%2B+log%28Y%2C8%29=2
log%28X%2C9%29%2B+log%288%2Cy%29=8%2F3
(log to the base 9 of X) + (log to the base Y of 8) = 2
(log to the base X of 9) + (log to the base 8 of Y) =8/3 or eight thirds
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
log9(x)+ logy(8) =2 ----(1)
logx(9)+log8(y) = 8/3 ----(2)
Put log9(x) = a----(3) and logy(8) = b----(4)
Putting (3) and (4) in (1), we get
a+b = 2 ----(5)
Using lomn(m) = 1/logm(n) in (2)
1/log9(x) +1/logy(8) = 8/3
That is 1/a +1/b = 8/3 ----(6) using (3) and (4)
Multiplying by 3ab through out,
3b+3a = 8ab
3(b+a) = 8ab
3X2 = 8ab ( using (5)and putting (a+b)= 2 )
Dividing by 2
3 = 4ab ----(7)
We know that by formula
(a-b)^2 = (a+b)^2 - 4ab
(a-b)^2 = (2)^2- 3
(using (5) and putting a+b = 2 and using (7) and putting 4ab= 3 )
(a-b)^2 = 4-3 = 1
(a-b)^2 = 1
(a-b) = 1 ----(8) (taking the positive root)
(a+b) = 2 ----(5) (writing (5) underneath (8) to make solving look easier)
(8) + (5) implies
(a+a) + 0 = (1+2)
2a=3
a=3/2
a=3/2 in (5) implies b = 1/2
Now a= 3/2 means
log9(x) = 3/2 (using (3) )
which implies x = (9)^(3/2)
(we shall retain the base 9 as it is the base in the problem)
[= (3^2)^(3/2) = (3)^3 = 27]
(using definition logb(N) =p implies and implied by N = (b)^p where N>0 )
And [(m)^n](p/q) = (m)^(np/q) )
That is x = 27
And Now b= 1/2 means
logy(8) = 1/2 (using (4) )
which implies 8 = (y)^(1/2)
(using definition logb(N) =p implies and implied by N = (b)^p where N>0 )
Squaring both the sides
8^2 = y
That is y = 8^2
Answer: x = [9^(3/2)] (= 27) and y = (8)^(2) (=64)
That is x = 27 and y = 64
Verification:We used (1) for finding b using a
Therefore we shall use (2) for verification:
logx(9)+log8(y) = 8/3 ----(2)
LHS = 1/log9(x) + log8(64)
=1/[log9[9^(3/2)]] + log8 [(8)^2]
=1/[(3/2)Xlog9(9)] + (2)Xlog8(8)
(since log(anything) to the base samething is =1)
=1/[(3/2)X1]+ 2X1
=1/[3/2]+ 2
=2/3+2
=(2+6)/3
=8/3 = RHS