Need help solving for x with work shown

2log6(4x)=0

Log2(x)+log2(x-3)=2

2log6(4x)=0

log6 (16 x^2) = 0

6^log6 (16 x^2) = 16x^2 = 6^0 = 1
so
x^2 = 1/16
x = 1/4
normally I would say also -1/4 but try to take the log of -1 :)

Log2(x)+log2(x-3)=2

2^[Log2(x)+log2(x-3)] =2^2

2^log2(x^2-3x) = 2^2

(x^2-3x) = 4

x^2 - 3 x - 4 = 0

(x-4)(x+1) = 0

x = 4 or x = -1

Sure! I can help you solve both equations and show the work.

Let's start with the first equation:

2log6(4x) = 0

To solve this equation, we need to isolate the logarithmic term and then apply the properties of logarithms.

Step 1: Divide both sides of the equation by 2 to cancel out the coefficient:
log6(4x) = 0 / 2
log6(4x) = 0

Step 2: Convert the logarithmic equation into exponential form. In general, if logb(x) = y, then by = x. Applying this to our equation, we have:
6^0 = 4x
1 = 4x

Step 3: Solve for x:
Divide both sides of the equation by 4:
1/4 = x

Therefore, the solution to the first equation is x = 1/4.

Now let's move on to the second equation:

log2(x) + log2(x-3) = 2

To solve this equation, we can simplify it using the properties of logarithms.

Step 1: Combine the logarithmic terms using the product rule of logarithms. According to the rule, logb(x) + logb(y) = logb(xy). Applying this to our equation, we get:
log2(x(x-3)) = 2

Step 2: Rewrite the equation in exponential form:
2^2 = x(x-3)
4 = x^2 - 3x

Step 3: Rearrange the equation to form a quadratic equation:
x^2 - 3x - 4 = 0

Step 4: Factorize the quadratic equation:
(x - 4)(x + 1) = 0

Step 5: Set each factor equal to zero and solve for x:
x - 4 = 0 or x + 1 = 0

Solving each equation separately:
x = 4 or x = -1

Therefore, the solutions to the second equation are x = 4 and x = -1.

I hope this helps! Let me know if you have any further questions.