Question

Hallar X
1.log₃(x+7)-log₃(2x)=0
2.log₄(x+4)-log₄(x-2)=1
3.log₅(x+20)-log(x-4)=2

Answer 1

1) log₃(x+7)-log₃(2x)=0

Logaritmo es biyectiva, entonces:

x+7 = 2x

x = 7

2) log₄(x+4)-log₄(x-2)=1

log₄[(x+4)/(x+2)] = 1

(x+4)/(x+2) = 4

x+4 = 4x + 8

x = - 4/3