math
posted by james .
Solve for x in the following equations. (4 marks)
log3(x  5) + log3(x + 3) = 2 < both log3.. 3 is subscript
log7x + log7(x  1) = log72x < all log7.. 7 is subscript.

use your log rules:
log (AB) = logA + logB
and
log (A/B) = logA  logB
1st one:
log_{3} [(x5)(x+3)] = 2
by definition of logs
(x5)(x+3) = 3^2 = 9
x^2  2x 24=0
(x6)(x+4) = 0
x = 6 or x=4 , but x=4 would make log_{3}(x+3) undefined, so
x = 6
2nd:
log_{7}(x) + log_{7}(x1)  log_{7}(2x) = 0
log_{7}[x(x1)/(2x)] = 0
log_{7}[(x1)/2] = 0
(x1)/2 = 7^0
(x1)/2 = 1
x1=2
x = 3