Solve for x in the equation:log_10 x+log _10 (x+2)=2
Using the logarithmic property $\log_a b + \log_a c = \log_a (b \cdot c)$, we can rewrite the equation as: $\log_{10} (x \cdot (x+2)) = 2$.
Since $\log_a b = c$ is equivalent to $a^c = b$, we have $10^2 = x \cdot (x+2)$.
Simplifying, we get $100 = x^2 + 2x$.
Moving all terms to one side, we have $x^2 + 2x - 100 = 0$.
Factoring or using the quadratic formula to solve this quadratic equation, we find that $x = -10$ or $x = 8$.
However, since $\log_{10} x$ requires $x$ to be positive, we discard the solution $x = -10$.
Therefore, the solution to the equation $\log_{10} x + \log_{10} (x+2) = 2$ is $x = 8$.
To solve the equation log_10 x + log_10 (x+2) = 2, we can use the properties of logarithms to simplify the equation before solving for x.
Using the property log(a) + log(b) = log(a*b), we can combine the two logarithms:
log_10(x(x+2)) = 2
Next, we can rewrite the logarithmic equation in exponential form:
10^2 = x(x+2)
Simplifying, we have:
100 = x(x+2)
Expanding the equation:
100 = x^2 + 2x
Rearranging the equation to quadratic form:
x^2 + 2x - 100 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Since this equation can't be easily factored, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation, a = 1, b = 2, and c = -100.
Plugging the values into the quadratic formula and simplifying, we have:
x = (-(2) ± √((2)^2 - 4(1)(-100))) / (2(1))
Simplifying further:
x = (-2 ± √(4 + 400)) / 2
x = (-2 ± √404) / 2
Now, we can simplify the expression under the square root:
x = (-2 ± √(4 * 101)) / 2
x = (-2 ± 2√101) / 2
Dividing both the numerator and the denominator by 2:
x = -1 ± √101
So, the solutions to the equation log_10 x + log_10 (x+2) = 2 are:
x = -1 + √101
and
x = -1 - √101