PRE CALCULUS- Exponential and Logarithmic Functions BRAINLIEST ANSWER *PLEASE ANSWER FAST* hey good luck you got it guys

Explanation:

We'll use these two log rules

So,

log(2x) + log(x-1) = log(6x)

log[ 2x*(x-1) ] = log(6x) use rule 1

2x*(x-1) = 6x use rule 2

2x^2-2x = 6x

2x^2-2x-6x = 0

2x^2-8x = 0

2x(x-4) = 0

2x = 0 or x-4 = 0

x = 0 or x = 4

The two possible solutions are x = 0 or x = 4. However, we must check both possible answers because some or all of them may be extraneous.

Checking x = 0

log(2x) + log(x-1) = log(6x)

log(2*0) + log(0-1) = log(6*0)

log(0) + log(-1) = log(0)

We run into a roadblock. We cannot take the log of 0 or anything negative. The domain of y = log(x) is x > 0. So we cross x = 0 off the list.

Now let's check x = 4

log(2x) + log(x-1) = log(6x)

log(2*4) + log(4-1) = log(6*4)

log(8) + log(3) = log(24)

log(8*3) = log(24)

log(24) = log(24)

We've confirmed x = 4. This is the only solution.

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step-by-step explanation: