Solve the following exponential equations: a) 2^x, b) 3^x, c) 10^x, d) 10^x

$$x^2 - 2x + 5 = 2^2$$ $$x^2 - 2x + 5 = 4$$

$$x^2 - 2x + 5 - 4 = 0$$ $$x^2 - 2x + 1 = 0$$

$$(x-1)^2 = 0$$ $$x-1 = 0$$ $$x = 1$$ $$\boxed{x = 1}$$

b) $\log_3 (x^2 + 2x + 2) = 0$

$$x^2 + 2x + 2 = 3^0$$ $$x^2 + 2x + 2 = 1$$

$$x^2 + 2x + 2 - 1 = 0$$ $$x^2 + 2x + 1 = 0$$

$$(x+1)^2 = 0$$ $$x+1 = 0$$ $$x = -1$$ $$\boxed{x = -1}$$

c) $\log_4 (x^2 + 15) = 3$

$$x^2 + 15 = 4^3$$ $$x^2 + 15 = 64$$

$$x^2 = 64 - 15$$ $$x^2 = 49$$

$$x = \pm\sqrt{49}$$ $$x = \pm 7$$ $$\boxed{x = \pm 7}$$

d) $\log_{10} (x^2 - 5x + 50) = 2$

$$x^2 - 5x + 50 = 10^2$$ $$x^2 - 5x + 50 = 100$$

$$x^2 - 5x + 50 - 100 = 0$$ $$x^2 - 5x - 50 = 0$$

$$(x-10)(x+5) = 0$$

This gives two solutions:

$$x-10 = 0 \implies x = 10$$ $$x+5 = 0 \implies x = -5$$ $$\boxed{x = 10, -5}$$

E) Solve the following

a) $\log_x 81 = 34$

$$x = 81^{1/34}$$

Since $81 = 3^4$:

$$x = (3^4)^{1/34}$$ $$x = 3^{4/34}$$ $$x = 3^{2/17}$$ $$\boxed{x = 3^{2/17}}$$

b) $\log_x 32 = 5$

Since $2^5 = 32$:

$$x^5 = 2^5$$ $$x = 2$$ $$\boxed{x = 2}$$

c) $\frac{1}{2} \log_x 64 = 6$

$$\log_x 64 = 6 \times 2$$ $$\log_x 64 = 12$$

Since $64 = 2^6$:

$$x^{12} = 2^6$$

Take the 12th root of both sides:

$$x = (2^6)^{1/12}$$ $$x = 2^{6/12}$$ $$x = 2^{1/2}$$ $$x = \sqrt{2}$$ $$\boxed{x = \sqrt{2}}$$

F) Change the following logarithmic form into index form
The general conversion from logarithmic form $\log_b M = N$ to index form is $b^N = M$.

a) $\log_2 128 = 7$

$$2^7 = 128$$ $$\boxed{2^7 = 128}$$

b) $\log_{10} (0.001) = -2$
Converting the given statement to index form:

$$10^{-2} = 0.001$$

(Note: While $10^{-2} = 0.01$, the instruction is to convert the given logarithmic form, not to verify its truth.)

$$\boxed{10^{-2} = 0.001}$$

c) $\log_7 49 = 2$

$$7^2 = 49$$ $$\boxed{7^2 = 49}$$