# 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 ->  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

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 Algebra: Logarithm Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on logarithm Question 29446: Solve the following system for (x,y): (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 thirdsAnswer by sdmmadam@yahoo.com(530)   (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