Solve log.
2 log (x-1)= log (x+1)
the answer is 3 but i don't know how to get it...Any one hint or suggestion will be greatly appreciated :)
From previous ans:
2 log (x-1)= log (x+1)
log[(x-1)²] = log(x+1)
Take antilog:
(x-1)²=x+1
Expand and solve quadratic:
x²-3x=0
x(x-3)=0
x=0 is not admissible as solution because log(-1) is undefined.
The remaining answer is therefore x=3.
To solve the equation: 2 log (x-1) = log (x+1)
Let's start by using the properties of logarithms.
1. Apply the product rule of logarithms: log(a) + log(b) = log(ab)
2 log (x-1) = log (x+1)
log((x-1)^2) = log (x+1)
2. Remove the logarithm on both sides by taking the antilogarithm or exponentiation. In this case, raise both sides of the equation to the base of the logarithm, which is 10.
(x-1)^2 = x + 1
3. Expand the equation and simplify:
x^2 - 2x + 1 = x + 1
4. Rearrange the equation to one side to isolate the variable:
x^2 - 2x - x + 1 - 1 = 0
x^2 - 3x = 0
5. Factor out the common factor in the equation:
x(x - 3) = 0
Now, set each factor equal to zero and solve for x:
1. x = 0
2. x - 3 = 0
x = 3
Therefore, the solution is x = 3.