Find the values of p in the equation log base 3 of p + 3 log base p of 3 = 4.

Tumia kanuni ya kubadilisha msingi wa logariti: $\log_b a = \frac{1}{\log_a b}$.
Hivyo, $\log_p 3 = \frac{1}{\log_3 p}$.
Badilisha hii kwenye mlinganyo asili:

$$\log_3 p + 3 \left(\frac{1}{\log_3 p}\right) = 4$$ $$\log_3 p + \frac{3}{\log_3 p} = 4$$

Mlinganyo unakuwa:

$$x + \frac{3}{x} = 4$$

Zidisha pande zote mbili kwa $x$ (kwa kuwa $x \neq 0$):

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

Panga upya mlinganyo kuwa fomu ya quadratic:

$$x^2 - 4x + 3 = 0$$

Faktorisha mlinganyo wa quadratic:

$$(x - 1)(x - 3) = 0$$

Hii inatoa suluhisho mbili zinazowezekana kwa $x$:

$$x = 1 \quad \text{au} \quad x = 3$$

Kesi 1: $x = 1$

$$\log_3 p = 1$$

Kutoka kwa ufafanuzi wa logariti ($b^c = a$ ikiwa $\log_b a = c$):

$$p = 3^1$$ $$p = 3$$

Kesi 2: $x = 3$

$$\log_3 p = 3$$

Kutoka kwa ufafanuzi wa logariti:

$$p = 3^3$$ $$p = 27$$

Kwa logariti $\log_b a$, msingi $b$ lazima uwe $b > 0$ na $b \neq 1$, na hoja $a$ lazima iwe $a > 0$.
Katika mlinganyo wetu, $p$ ni hoja ya $\log_3 p$ na msingi wa $\log_p 3$.
Hivyo, $p$ lazima iwe $p > 0$ na $p \neq 1$.
Maadili yote mawili $p=3$ na $p=27$ yanakidhi masharti haya.

Kwa $p=3$: $\log_3 3 + 3 \log_3 3 = 1 + 3(1) = 4$. (Sahihi)
Kwa $p=27$: $\log_3 27 + 3 \log_{27} 3 = 3 + 3(\frac{1}{3}) = 3 + 1 = 4$. (Sahihi)

Maadili ya $p$ ni:

$$p = 3 \quad \text{au} \quad p = 27$$