Find the exact value for x in the following equation log2(x+4)=2−log2(1/8)

Add $\log_2 \frac{1}{8}$ to both sides:

$$\log_2(x+4) + \log_2 \frac{1}{8} = 2$$

We know that $\frac{1}{8} = 2^{-3}$.

$$\log_2 \frac{1}{8} = \log_2 (2^{-3}) = -3$$

Substitute this value back into the equation:

$$\log_2(x+4) + (-3) = 2$$ $$\log_2(x+4) - 3 = 2$$

Add 3 to both sides:

$$\log_2(x+4) = 2 + 3$$ $$\log_2(x+4) = 5$$

The definition of a logarithm states that if $\log_b A = C$, then $b^C = A$.
Here, $b=2$, $A=(x+4)$, and $C=5$.

$$2^5 = x+4$$

Calculate $2^5$:

$$32 = x+4$$

Subtract 4 from both sides:

$$x = 32 - 4$$ $$x = 28$$

For $\log_2(x+4)$ to be defined, $x+4 > 0$.
If $x=28$, then $28+4 = 32 > 0$, which is valid.

The final answer is $\boxed{28}$.

2 zimekamilika. Zimebaki 3 leo — tuma inayofuata.