Question

log3 (x-2) + log 3 (x+4)= log3 27

e

log2 (x-2) + log2 (x-3) = 1

Answer 1

log3 (x-2) + log 3 (x+4)= log3 27

(x-2)•(x+4)=27

x² +2x -8 - 27 =0

x² +2x -35= 0

∆=2² -4.1.(-35) =144 => √∆=12

x' = -2+12/2 = 10/2 = 5 ✓

x" = -2-12/2= -14/2= -7 (não serve)

log2 (x-2) + log2 (x-3) = 1

log2 (x-2) + log2 (x-3) = log2 2

(x-2)•(x-3) = 2

x² -5x +6 - 2 = 0

x² -5x +4 = 0

(x-4) . (x-1) = 0

x'=4 ✓..x"=1 (não Serve)