Are you allowed to do this
2log(x-1) =log(x+1)
2 (x-1)=(x+1)
x=3
Please help :)
See:
http://www.jiskha.com/display.cgi?id=1294279972
Some mathteacher published a book some time ago listing all the silly solutions that students had produced.
It actually had it this way
2log(x-1) =log(x+1)
(2log(x-1))/log = (log(x+1))/log , then the log/log canceled
2(x-1) = (x+1)
2x - 2 = x+1
x = 3, which by sheer coincidence is the right answer.
My teacher showed us this one:
64/16
= (6)4/1(6) [cancel the 6]
= 4/1
= 4
gets the right answer, but the wrong way.
Yes, I am allowed to help you with this question. Let's go through the steps together to solve the equation.
Step 1: Start with the given equation: 2log(x-1) = log(x+1)
Step 2: Use the logarithmic rule that states log(a) - log(b) = log(a/b) to simplify the equation. Apply this rule to the left side of the equation:
log((x-1)^2) = log(x+1)
Step 3: Apply the rule that if log(a) = log(b), then a = b. Set the expressions inside the logarithms equal to each other:
(x-1)^2 = x+1
Step 4: Expand and simplify the equation:
x^2 - 2x + 1 = x + 1
Step 5: Combine like terms:
x^2 - 2x - x = -1 - 1
x^2 - 3x = -2
Step 6: Move all terms to one side to create a quadratic equation:
x^2 - 3x + 2 = 0
Step 7: Factor the quadratic equation to solve for x:
(x - 1)(x - 2) = 0
Using the zero product property, we set each factor equal to zero:
x - 1 = 0 --> x = 1
x - 2 = 0 --> x = 2
Therefore, the solution to the original equation is x = 1 or x = 2.