Solve the following simultaneous equations (a) 2^x = 8*2^x / log(2x−3y)=1 (b) log(y+3)−1 = log x / log (x+4)+log (x−4) = log (y-2)

Fomula ni $P(t) = P_0 e^{rt}$.
Tunaambiwa:

• Idadi ya awali ($P_0$) = 200
• Idadi huongezeka mara mbili katika masaa 5. Hii inamaanisha kuwa $P(5) = 2 \times P_0 = 2 \times 200 = 400$ wakati $t = 5$ masaa.

Tunajua $P(5) = 400$, $P_0 = 200$, na $t = 5$.

$$400 = 200 e^{r \times 5}$$

Gawanya pande zote mbili kwa 200:

$$\frac{400}{200} = e^{5r}$$ $$2 = e^{5r}$$

Chukua ln ya pande zote mbili:

$$\ln(2) = \ln(e^{5r})$$

Kutokana na sheria za logarithm, $\ln(e^x) = x$:

$$\ln(2) = 5r$$

Gawanya pande zote mbili kwa 5:

$$r = \frac{\ln(2)}{5}$$

Tumia kikokotoo kupata thamani ya $\ln(2) \approx 0.6931$:

$$r = \frac{0.6931}{5}$$ $$r \approx 0.13862$$

Kiwango cha ukuaji $r$ ni takriban 0.1386 (kwa sehemu nne za desimali).

$$P(t) = P_0 e^{rt}$$ $$400 = 200 e^{r \times 5}$$ $$2 = e^{5r}$$ $$\ln(2) = 5r$$ $$r = \frac{\ln(2)}{5}$$ $$r \approx \frac{0.693147}{5}$$ $$r \approx 0.138629$$

Kiwango cha ukuaji $r$ ni $\boxed{0.1386}$.