The problem asks to evaluate the expression $\log_8 256 + \log_8 4 - \log_8 7 + \log_8 224$ using the laws of logarithms.

1. Problem Description

The problem asks to evaluate the expression $\log_8 256 + \log_8 4 - \log_8 7 + \log_8 224$ using the laws of logarithms.

2. Solution Steps

We can use the properties of logarithms to simplify the expression.

First, we can use the property $\log_a x + \log_a y = \log_a(xy)$ to combine the first two terms and the last two terms:

$\log_8 256 + \log_8 4 = \log_8(256 \cdot 4) = \log_8 1024$

$-\log_8 7 + \log_8 224 = \log_8(\frac{224}{7}) = \log_8 32$

So the original expression becomes:

$\log_8 1024 + \log_8 32$

Using the same property again, we get:

$\log_8 (1024 \cdot 32) = \log_8 32768$

Now, we need to find $x$ such that $8^x = 32768$.

Since $8 = 2^3$, we can rewrite the equation as $(2^3)^x = 32768$.

We can also write $32768$ as $2^{15}$.

Therefore, $(2^3)^x = 2^{15}$, which means $2^{3x} = 2^{15}$.

Equating the exponents, we have $3x = 15$, so $x = 5$.

Therefore, $\log_8 32768 = 5$.

Alternatively, we can evaluate each term separately.

$\log_8 256 = \log_8 (8^{8/3}) = \log_8 (2^8) = \frac{8}{3} \times \frac{\log 2}{\log 2^3}=8/3$ since $256 = 2^8 = (2^3)^{8/3} = 8^{8/3}$. But $256 = 4*64 = 4*8^2$, so we have $\log_8 256 = \log_8(8^2 * 4) = 2 + \log_8 4 = 2 + \log_8 2^2 = 2 + \frac{2}{3} = 8/3 = log_8 (2^8) = log_8 (8^{8/3})$.

Also, $\log_8 4 = \log_8 2^2 = 2 \log_8 2 = 2(\frac{1}{3}) = \frac{2}{3}$.

$\log_8 7$ can't be easily simplified.

$\log_8 224 = \log_8 (32 \cdot 7) = \log_8 32 + \log_8 7 = \log_8 2^5 + \log_8 7 = \frac{5}{3} + \log_8 7$.

Therefore, $\log_8 256 + \log_8 4 - \log_8 7 + \log_8 224 = \frac{8}{3} + \frac{2}{3} - \log_8 7 + \frac{5}{3} + \log_8 7 = \frac{8+2+5}{3} = \frac{15}{3} = 5$.

3. Final Answer

5