Log5(x=1) - 1 = 1+ log5(x-1); solve

Question

My daughter is a student fo Class Nine, I have not solved it. Please solve as early as possible

Answer 1

me suspekt yu tried tu tipe

log5(x-1) -1=1+log5(x-1)

but that become 0=2

??? maebee it shood be log5(x+1)=2+log5(x-1)????

or log5(x+1)-log5(x-1)=2

or log5[(x+1) / (x-1)]=2

(x+1) / (x-1)=5^2=25

x+1=25*(x-1)=25x -25

25x-x=25+1

x=26/24=1.0833333333...

Answer 2

Did you mean log5(x-1) - 1 = 1 + log5(x-1) ?

Subtract log5(x-1) from both sides and you get:

-1 = 1

That can't happen, so the equation is wrong / unsolvable.

 

Did you mean log5(x+1) - 1 = 1 + log5(x-1) ?

Subtract log5(x-1) from both sides:

log5(x+1) - log5(x-1) - 1 = 1

A property of logs is log(a) - log(b) = log(a/b)

log5( (x+1)/(x-1) ) -1 = 1

Add 1 to both sides:

log5( (x+1)/(x-1) ) = 2

Another property of logs is that loga(b) = c means a^c = b

5^2 = (x+1)/(x-1)

25 = (x+1)/(x-1)

25(x-1) = x+1

25x - 25 = x + 1

24x - 25 = 1

24x = 26

x = 26/24

x = 13/12