F.4 Number and Algebra

1) log8x2/log2x

= (log x^2 / log 8) / (log x / log 2)

= (2log x / log x)(log 2 / log 8)

= 2(log2 / log8)

= 2 (log2 / 3log2)

= 2/3

2)log5(3x-1)-log5 (x+1)-1 = 0

log5[(3x-1)/(x+1)] = 1

log5 [(3x-1)/(x+1)] = log5 5

(3x-1) / (x+1) = 5

3x-1 = 5x+5

2x = - 6

x = - 3

It becomes
log5(3(-3)-1)-log5(-3+1)-1=0

No solution since negative number
cannot be taken log.

1. 2/3~~~~~~~~~~~~