Posted by Kong on .
solve it:
a) 3 log x with base 8 + 2 = 0
b)2 log (x1)with base 6 =log 4 with base 6

Maths(logarithmic functions) 
Reiny,
a)
3 log_{8} x = 2
log_{8} x = 2/3
8^(2/3) = x
x = 1/4
b) 2 log_{6} (x1) = log_{6} 4
log_{6} (x1)^2 = log_{6} 4
(x1)^2 = 4
x1 = ±2
x = 3 or x = 1 , but for log(x1) to be defined, x>1
so x = 3