Question

1) ³log(2x-5) =3
2)7log(x²+3x+5) =7log5
3) log(x²+4x+5) =1

please dg cara penyelesaianya..

Answer 1

1. ³log (2x - 5) = ³log3³
(2x - 5) = 3³
2x - 5 = 27
2x = 32
x = 16

2. 7log(x²+3x+5) = 7log5
(x²+3x+5) = 5
x²+3x = 0
x ( x+3) = 0
x = 0 atau x = -3

3. log (x²+4x+5 ) = 1
log (x²+4x+5)=log10^1
(x²+4x+5) = 10^1
x² + 4x - 5 = 0
(x - 1) (x + 5) = 0
x = 1 atau x = -5

semoga membantu :)

Answer 2

1) 3log (2x-5) =3
3log (2.16-5) =3
3log (32-5) =3
3log 27 =3