Solve log2(7x-3)>= log2(x+12)
My work:
6x=15
x=15/6
x-5/2
so my answer would be: x>= 5/2
Is this correct?
I think you are right
To solve the inequality log2(7x-3) >= log2(x+12), you need to follow these steps:
Step 1: Apply the property of logarithms that states log(base a)(b) >= log(base a)(c) if and only if b >= c.
Step 2: Set the expressions inside the logarithms equal to each other and solve for x.
log2(7x-3) = log2(x+12)
Step 3: Since the bases of the logarithms are the same (base 2), you can remove the logarithms and set the expressions inside them equal to each other:
7x-3 = x+12
Step 4: Simplify and solve for x:
Subtract x from both sides:
7x - x - 3 = x - x + 12
6x - 3 = 12
Add 3 to both sides:
6x - 3 + 3 = 12 + 3
6x = 15
Divide both sides by 6:
(6x)/6 = 15/6
x = 15/6 = 2.5
So, the solution to the equation log2(7x-3) >= log2(x+12) is x >= 2.5.