Solve the following logarithmic equations: 1. log(x²−4x+7)=2 2. log10(x²−3x+12)=1 3. log3(x²+12x+27)=2 4. log4(x²+6x+11)¹/2 5. log5(x³−9)=0

$$5^N = \frac{1}{5^3}$$ $$5^N = 5^{-3}$$

$$N = -3$$

So, the logarithmic form is $\log_5 (1/125) = -3$.
Converting to index form:

$$\boxed{5^{-3} = \frac{1}{125}}$$

G) Solve the following logarithmic equations

1) $\log (x^2 - 4x + 7) = 2$
Assuming the base is 10 (common logarithm).

$$x^2 - 4x + 7 = 10^2$$ $$x^2 - 4x + 7 = 100$$

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-93)}}{2(1)}$$ $$x = \frac{4 \pm \sqrt{16 + 372}}{2}$$ $$x = \frac{4 \pm \sqrt{388}}{2}$$ $$x = \frac{4 \pm 2\sqrt{97}}{2}$$ $$x = 2 \pm \sqrt{97}$$

The expression $x^2 - 4x + 7$ is always positive, so both solutions are valid.

$$\boxed{x = 2 \pm \sqrt{97}}$$

2) $\log_{10} (x^2 - 3x + 12) = 1$

$$x^2 - 3x + 12 = 10^1$$ $$x^2 - 3x + 12 = 10$$

$$(x-1)(x-2) = 0$$ $$x = 1 \quad \text{or} \quad x = 2$$

The expression $x^2 - 3x + 12$ is always positive, so both solutions are valid.

$$\boxed{x = 1, 2}$$

3) $\log_3 (x^2 + 12x + 27) = 2$

$$x^2 + 12x + 27 = 3^2$$ $$x^2 + 12x + 27 = 9$$

$$x = \frac{-12 \pm \sqrt{12^2 - 4(1)(18)}}{2(1)}$$ $$x = \frac{-12 \pm \sqrt{144 - 72}}{2}$$ $$x = \frac{-12 \pm \sqrt{72}}{2}$$ $$x = \frac{-12 \pm 6\sqrt{2}}{2}$$ $$x = -6 \pm 3\sqrt{2}$$

Both solutions satisfy the domain condition $x^2 + 12x + 27 > 0$.

$$\boxed{x = -6 \pm 3\sqrt{2}}$$

4) $\log_x (x^2 + 6x + 11) = 1/2$

$$x^{1/2} = x^2 + 6x + 11$$ $$\sqrt{x} = x^2 + 6x + 11$$

For the logarithm to be defined, the base $x$ must be $x > 0$ and $x \neq 1$. Also, $x^2 + 6x + 11 > 0$. The discriminant of $x^2 + 6x + 11$ is $6^2 - 4(1)(11) = 36 - 44 = -8 < 0$. Since the leading coefficient is positive, $x^2 + 6x + 11$ is always positive.

Let $f(x) = x^2 + 6x + 11 - \sqrt{x}$. We are looking for $f(x)=0$.
The minimum value of $x^2 + 6x + 11$ occurs at $x = -6/(2 \times 1) = -3$. At this point, $x^2 + 6x + 11 = (-3)^2 + 6(-3) + 11 = 9 - 18 + 11 = 2$.
Since $x > 0$, the minimum value of $x^2 + 6x + 11$ for $x>0$ is greater than 2 (e.g., at $x=0$, it's 11; at $x=1$, it's 18).
The function $\sqrt{x}$ for $x>0$ has values $\sqrt{x} \ge 0$.
For $\sqrt{x} = x^2 + 6x + 11$ to have a solution, $\sqrt{x}$ must be at least 2, which means $x \ge 4$.
If $x=4$, $\sqrt{4}=2$, but $4^2+6(4)+11 = 16+24+11 = 51$. So $2 \neq 51$.
As $x$ increases, $x^2 + 6x + 11$ grows much faster than $\sqrt{x}$.
Therefore, there are no real values of $x$ that satisfy this equation.

$$\boxed{\text{No real solutions}}$$

5) $\log_5 (x^3 - 9) = 0$

$$x^3 - 9 = 5^0$$ $$x^3 - 9 = 1$$

$$x^3 = 10$$ $$x = \sqrt[3]{10}$$

For the logarithm to be defined, $x^3 - 9 > 0$, which means $x^3 > 9$, or $x > \sqrt[3]{9}$.
Since $\sqrt[3]{10} > \sqrt[3]{9}$, the solution is valid.

$$\boxed{x = \sqrt[3]{10}}$$

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