Solve for k given that (log3 k)2 - log3 k0.5 = log2 8

Tunatumia sheria ya logariti $\log_b (x^p) = p \log_b x$ kurahisisha $\log_3 k^{0.5}$:

$$\log_3 k^{0.5} = 0.5 \log_3 k$$

Kisha, tunatathmini $\log_2 8$:
Kwa kuwa $2^3 = 8$, basi $\log_2 8 = 3$.

Mlinganyo unakuwa:

$$(\log_3 k)^2 - 0.5 \log_3 k = 3$$

$$x^2 - 0.5x = 3$$

Panga upya mlinganyo katika fomu ya kawaida ya quadratic $ax^2 + bx + c = 0$:

$$x^2 - 0.5x - 3 = 0$$

Ili kuepuka desimali, zidisha mlinganyo mzima kwa 2:

$$2x^2 - x - 6 = 0$$

Tutatumia njia ya kuzidisha (factoring):

$$2x^2 - 4x + 3x - 6 = 0$$ $$2x(x - 2) + 3(x - 2) = 0$$ $$(2x + 3)(x - 2) = 0$$

Hii inatoa thamani mbili zinazowezekana kwa $x$:

$$2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}$$ $$x - 2 = 0 \implies x = 2$$

Kesi 1: $x = 2$

$$\log_3 k = 2$$

Kwa ufafanuzi wa logariti, $k = 3^2$.

$$k = 9$$

Kesi 2: $x = -\frac{3}{2}$

$$\log_3 k = -\frac{3}{2}$$ $$k = 3^{-\frac{3}{2}}$$ $$k = \frac{1}{3^{3/2}}$$ $$k = \frac{1}{\sqrt{3^3}}$$ $$k = \frac{1}{\sqrt{27}}$$ $$k = \frac{1}{3\sqrt{3}}$$

Ili kurahisisha denominator, zidisha kwa $\frac{\sqrt{3}}{\sqrt{3}}$:

$$k = \frac{1}{3\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3 \times 3} = \frac{\sqrt{3}}{9}$$

Thamani zote mbili za $k$ ni chanya, hivyo zote ni suluhisho halali.

Majibu ya $k$ ni:

$$k = 9 \text{ au } k = \frac{\sqrt{3}}{9}$$