Lg(2x+2)+lg(15-x)=1+lg3

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To solve this logarithmic equation, we can start by simplifying the logarithms on the left side of the equation using the properties of logarithms.

Using the property log(a) + log(b) = log(ab), we can rewrite the equation as:

lg[(2x+2)(15-x)] = 1 + lg(3)

Next, we can expand the expression inside the logarithm:

lg(30x - 2x^2 + 30 - 2) = 1 + lg(3)

Now, using the property log(a) = b is equivalent to a = 10^b, we can write the equation in exponential form:

30x - 2x^2 + 30 - 2 = 10^(1) * 3

30x - 2x^2 + 28 = 3

Rearranging this equation, we get a quadratic equation:

-2x^2 + 30x + 28 = 3

-2x^2 + 30x + 25 = 0

Now, we can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values of a=-2, b=30, and c=25 into the formula, we get:

x = (-30 ± √(30^2 - 4(-2)25)) / 2*(-2)

x = (-30 ± √(900 + 200)) / -4

x = (-30 ± √1100) / -4

x = (-30 ± 10√11) / -4

Therefore, the solutions to the equation are:

x = (-30 + 10√11) / -4, and x = (-30 - 10√11) / -4

These are the two possible values of x that satisfy the given logarithmic equation.