Question 60838
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!!!