The problem requires finding the value of x that satisfies the given logarithmic equation:
$ \log x + \log (x - 7) = \log (x + 11) + \log 2 $
Use the logarithm property $ \log a + \log b = \log (ab) $ to simplify both sides of the equation.
The equation becomes:
$ \log (x(x - 7)) = \log (2(x + 11)) $
Since the logarithms are equal, their arguments must be equal:
$ x(x - 7) = 2(x + 11) $
Expand both sides:
$ x^2 - 7x = 2x + 22 $
Rearrange into a standard quadratic equation ($ ax^2 + bx + c = 0 $):
$ x^2 - 7x - 2x - 22 = 0 $
$ x^2 - 9x - 22 = 0 $
Factor the quadratic equation:
$ (x - 11)(x + 2) = 0 $
This gives two potential solutions: $ x = 11 $ or $ x = -2 $.
The argument of a logarithm must always be positive.
To satisfy all conditions, x must be greater than 7 ($ x > 7 $).
Evaluate the potential solutions against the domain requirement ($ x > 7 $):
Therefore, the only value of x that satisfies the original equation is 11.
Consider two distinct positive real numbers $m, n$, with $m > n$.
Let $x = n^{\log_{10}(m)}$ and $y = m^{\log_{10}(n)}$. The relation between $x$ and $y$ is _______.
For a real number $x > 1$,
$\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} = 1$
The value of $x$ is
Let $a = 30!$, $b = 50!$, and $c = 100!$. Consider the following numbers:
$log_a c$, $log_c a$, $log_b a$, $log_a b$
Which one of the following inequalities is CORRECT?