Solve the following simultaneous eqns:- 2 ln y = 3 ln 2 + ln x, y ln 3 = x ln 81

Tumia sheria za logariti $a \ln b = \ln (b^a)$ na $\ln a + \ln b = \ln (ab)$.

$$2 \ln y = 3 \ln 2 + \ln x$$ $$\ln (y^2) = \ln (2^3) + \ln x$$ $$\ln (y^2) = \ln 8 + \ln x$$ $$\ln (y^2) = \ln (8x)$$

Kwa kuwa $\ln A = \ln B$ inamaanisha $A = B$:

$$y^2 = 8x \quad \text{(Mlinganyo 1A)}$$

Tumia sheria ya logariti $\ln (b^a) = a \ln b$. Tunajua kuwa $81 = 3^4$.

$$y \ln 3 = x \ln 81$$ $$y \ln 3 = x \ln (3^4)$$ $$y \ln 3 = x (4 \ln 3)$$

Gawanya pande zote mbili kwa $\ln 3$ (kwa kuwa $\ln 3 \neq 0$):

$$y = 4x \quad \text{(Mlinganyo 2A)}$$

Badilisha $y = 4x$ kwenye $y^2 = 8x$:

$$(4x)^2 = 8x$$ $$16x^2 = 8x$$

Panga upya mlinganyo na utoe $x$:

$$16x^2 - 8x = 0$$ $$8x(2x - 1) = 0$$

Hii inatoa suluhisho mbili zinazowezekana kwa $x$:

$$8x = 0 \implies x = 0$$ $$2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$$

Kwa kuwa $\ln x$ imejumuishwa katika mlinganyo asili, $x$ lazima iwe kubwa kuliko 0. Kwa hiyo, $x=0$ si suluhisho halali.
Hivyo, $x = \frac{1}{2}$.

Tumia Mlinganyo 2A, $y = 4x$, na badilisha thamani ya $x = \frac{1}{2}$:

$$y = 4 \left(\frac{1}{2}\right)$$ $$y = 2$$

Kwa kuwa $\ln y$ imejumuishwa katika mlinganyo asili, $y$ lazima iwe kubwa kuliko 0. Kwa hiyo, $y=2$ ni suluhisho halali.

Suluhisho la milinganyo ni:

$$x = \frac{1}{2}, y = 2$$