Lg(x+6)-0.5lg(2x-3)=2-lg25

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To solve this equation, we need to simplify the logarithms first.

Given:
lg(x+6) - 0.5lg(2x-3) = 2 - lg(25)

Apply the properties of logarithms:
lg(x+6) - lg((2x-3)^0.5) = 2 - lg25
lg(x+6) - lg√(2x-3) = 2 - lg25

Now, use the property that lg√a = 0.5lg(a):
lg(x+6) - 0.5lg(2x-3) = 2 - lg25

Now, we can rewrite the equation without any square roots:
lg(x + 6) - 0.5lg(2x - 3) = 2 - lg(25)

Next, we can apply the properties of logarithms to simplify the equation further. We will combine the logarithms:
lg[(x + 6)/(√(2x - 3))] = 2 - lg(25)

Next, we can rewrite the right side in terms of lg(25):
lg[(x + 6)/(√(2x - 3))] = lg(100) - lg(25)

Now, we can simplify the right side to get a single logarithm:
lg[(x + 6)/(√(2x - 3))] = lg(4)

At this point, we have a single logarithm on each side. We can set the arguments of the logarithms equal to each other:
(x + 6)/(√(2x - 3)) = 4

Now, we can solve for x by cross multiplying:
4(√(2x - 3)) = x + 6

Next, we will square both sides of the equation to get rid of the square root:
16(2x - 3) = (x + 6)^2

Expand and simplify the equation:
32x - 48 = x^2 + 12x + 36

Rearrange the equation to set it equal to zero:
x^2 - 20x - 84 = 0

Now, we have a quadratic equation that we can solve using factoring, the quadratic formula, or other methods to find the values of x. Once we find the values of x, we can substitute them back into the original equation to check for extraneous solutions and verify the solution.