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.