SOLUTION: Would someone help me. Solve the following equations. Round to 4 decimals when necessary. a. logx 729/4096=6 b. log5 325=x c. log2 (x-2)+log2 (x+1)=2 d. log (2x+6)-log(x-

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Would someone help me. Solve the following equations. Round to 4 decimals when necessary. a. logx 729/4096=6 b. log5 325=x c. log2 (x-2)+log2 (x+1)=2 d. log (2x+6)-log(x-      Log On


   

Question 60838: Would someone help me.
Solve the following equations. Round to 4 decimals when necessary.
a. logx 729/4096=6
b. log5 325=x
c. log2 (x-2)+log2 (x+1)=2
d. log (2x+6)-log(x-1)=1
Thank you for your help!
COULD SOMEONE PLEASE HELP!!!
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Would someone help me.
Solve the following equations. Round to 4 decimals when necessary.
a. logx 729/4096=6
X^6 = 729/4096 = 3^6/4^6 = (3/4)^6
X=3/4
b. log5 325=x
5^X = 325 ..IT GIVES IRRATIOAL SOLUTION...CHEK THE PROBLEM..IS IT 625..IF O
5^X =625 = 5^4
X=4
c. log2 (x-2)+log2 (x+1)=2
LOG[(X-2)(X+1)] TO BASE 2 = 2
(X-2)(X+1)=2^2=4
X^2-X-2-4=0
X^2-3X+2X-6=0
X(X-3)+2(X-3)=0
(X-3)(X+2)=0
X=3 ....SINCE X=-2 LEADS TO LOG OF NEGATIVE NUMBER WHICH IS IMAGINARY.
d. log (2x+6)-log(x-1)=1
LOG[(2X+6)/(X-1)]=1
ASSUMING BASE TO BE 10
(2X+6)/(X-1)=10
2X+6 = 10X-10
8X = 16
X = 2


Thank you for your help!
COULD SOMEONE PLEASE HELP!!!