Evaluate the following logarithmic expressions: 7) 2 5 - 3 2 + 32 8) ( 32 + 128 - 729)/( 32 + 2 - 27)

$$2\log 5 = \log 5^2 = \log 25$$ $$3\log 2 = \log 2^3 = \log 8$$

The expression becomes:

$$\log 25 - \log 8 + \log 32$$

Combine the terms:

$$\log \left(\frac{25}{8}\right) + \log 32$$ $$\log \left(\frac{25}{8} \times 32\right)$$

Simplify the expression inside the logarithm:

$$\log (25 \times 4)$$ $$\log 100$$

Since the base is 10, $\log_{10} 100$ asks "10 to what power equals 100?".

$$\log 100 = 2$$

The value is $\boxed{2}$.

Identify powers of common bases: $32 = 2^5$, $128 = 2^7$, $729 = 3^6$.

$$\text{Numerator} = \log 2^5 + \log 2^7 - \log 3^6$$

Apply the power rule $a \log b = \log b^a$:

$$\text{Numerator} = 5\log 2 + 7\log 2 - 6\log 3$$

Combine like terms:

$$\text{Numerator} = 12\log 2 - 6\log 3$$

Identify powers of common bases: $32 = 2^5$, $27 = 3^3$.

$$\text{Denominator} = \log 2^5 + \log 2 - \log 3^3$$

Apply the power rule:

$$\text{Denominator} = 5\log 2 + \log 2 - 3\log 3$$

Combine like terms:

$$\text{Denominator} = 6\log 2 - 3\log 3$$

Substitute the simplified numerator and denominator back into the fraction:

$$\frac{12\log 2 - 6\log 3}{6\log 2 - 3\log 3}$$

Factor out common terms from the numerator and denominator:

$$\frac{6(2\log 2 - \log 3)}{3(2\log 2 - \log 3)}$$

Cancel out the common factor $(2\log 2 - \log 3)$:

$$\frac{6}{3} = 2$$

The value is $\boxed{2}$.

Since $\log A = \log B$ implies $A=B$:

$$x+5 = \frac{4}{x+2}$$

Multiply both sides by $(x+2)$:

$$(x+5)(x+2) = 4$$

Expand the left side:

$$x^2 + 2x + 5x + 10 = 4$$ $$x^2 + 7x + 10 = 4$$

Subtract 4 from both sides to set the quadratic equation to zero:

$$x^2 + 7x + 6 = 0$$

Find two numbers that multiply to 6 and add to 7 (which are 1 and 6):

$$(x+1)(x+6) = 0$$

This gives two potential solutions for $x$:

$$x+1 = 0 \implies x = -1$$ $$x+6 = 0 \implies x = -6$$

The argument of a logarithm must be positive.
For $\log(x+5)$, we need $x+5 > 0 \implies x > -5$.
For $\log(x+2)$, we need $x+2 > 0 \implies x > -2$.
Both conditions must be satisfied, so $x$ must be greater than $-2$.

Check $x = -1$:
$x+5 = -1+5 = 4 > 0$ (Valid)
$x+2 = -1+2 = 1 > 0$ (Valid)
So, $x = -1$ is a valid solution.

Check $x = -6$:
$x+5 = -6+5 = -1$ (Not valid, as the argument must be positive)
So, $x = -6$ is not a valid solution.

The value of $x$ that satisfies the equation is $\boxed{-1}$.