1. Solve log (4x) = log (2) + log (x-1)
the fact that this is x-1 and not + is really messing me up because i keep getting a negative number and that isnt possible
2. Solve: absolut value of(x^2 -4x-4) = 8
thanks so much!!! I have been studying and my final is tomorrow. these are just two questions that came up:)
log [ 4x/(2(x-1))] = 0
so
10^log [ 4x/(2(x-1))] = 10^0 = 1
so
4x/[2(x-1))] = 1
4x = 2 x - 2
2 x = - 2
x = -1
x^2 - 4 x - 4 = 8
x^2 - 4 x - 12 = 0
solve quadratic
then other possible solution
x^2 - 4 x -4 = -8
x^2 - 4 x + 4 = 0
(x-2)(x-2) = 0
x = 2
1. I disagree with Damon's solution.
log(4x) = log2 + log(x-1) , so x>1
log(4x) - log(x-1) = log2
log( 4x/(x-1) ) = log2
4x/(x-1) = 2
4x = 2x-2
2x = -2
x = -1 , but x > 1 or else log (4x) and log(x-1) are both undefined
no solution
thanks so much both of you!! Yah I wasn't sure about the first one but now I know how to do both!!
thanks!!
1. To solve the equation log (4x) = log (2) + log (x-1), we can use the properties of logarithms.
First, we can combine the logarithms on the right side using the product rule of logarithms, which states that log(a) + log(b) = log(ab). Therefore, log (2) + log (x-1) can be simplified to log (2(x-1)).
Now, we have the equation log (4x) = log (2(x-1)). According to the rule of logarithms, if two logarithms with the same base are equal, then the arguments of the logarithms must be equal as well.
So, we can set 4x = 2(x-1) and solve for x:
4x = 2x - 2
2x = -2
x = -1
However, you mentioned that you keep getting a negative number and that isn't possible. This indicates that there might be a mistake in your calculations.
2. To solve the equation |x^2 - 4x - 4| = 8, we can consider two cases: when the expression inside the absolute value is positive and when it is negative. Let's solve each case separately:
Case 1: x^2 - 4x - 4 is positive:
In this case, we can remove the absolute value and rewrite the equation as:
x^2 - 4x - 4 = 8
Rearranging the equation, we get:
x^2 - 4x - 12 = 0
Using factoring or the quadratic formula, we find two possible solutions:
x = 6 or x = -2
Case 2: x^2 - 4x - 4 is negative:
In this case, we need to negate the expression inside the absolute value and rewrite the equation as:
-(x^2 - 4x - 4) = 8
Distributing the negative sign, we get:
-x^2 + 4x + 4 = 8
Rearranging the equation, we have:
-x^2 + 4x - 4 = 0
Using factoring or the quadratic formula again, we find two more possible solutions:
x = 2 + √8 or x = 2 - √8
So, the solutions to the equation |x^2 - 4x - 4| = 8 are:
x = 6, x = -2, x = 2 + √8, and x = 2 - √8.
I hope this helps, and good luck with your final exam!