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(A) Let $P$ be the principal at time $t$. Given that the principal increases continuously at a rate of $x \%$ per year,we have the differential equati

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$6.93$

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(A) Let $P$ be the principal at time $t$. Given that the principal increases continuously at a rate of $x \%$ per year,we have the differential equation: $\frac{dP}{dt} = \frac{x}{100} P$. Integrating this,we get $\int \frac{dP}{P} = \int \frac{x}{100} dt$,which gives $\ln P = \frac{x}{100} t + C$. At $t = 0$,$P = P_0 = 100$. Thus,$C = \ln 100$. So,$\ln P = \frac{x}{100} t + \ln 100$,or $\ln(\frac{P}{100}) = \frac{x}{100} t$. Given that the principal doubles in $10$ years,at $t = 10$,$P = 200$. Substituting these values: $\ln(\frac{200}{100}) = \frac{x}{100} 	imes 10$. $\ln 2 = \frac{x}{10}$. Since $\ln 2 = 2.3026 	imes \log_{10} 2$,we have $\ln 2 = 2.3026 	imes 0.6931 \approx 1.596$. However,if the rate is defined such that $\frac{dP}{dt} = rP$ where $r = \frac{x}{100}$,then $P(t) = P_0 e^{rt}$. $200 = 100 e^{10r} \implies 2 = e^{10r} \implies 10r = \ln 2$. Using $\ln 2 \approx 0.6931$,$10r = 0.6931 \implies r = 0.06931$. Thus,$x = 100r = 6.931 \% \approx 6.93 \%$.

In a bank,the principal increases continuously at a rate of $x \%$ per year. Then the rate,$x$,if ₹$100$ doubles itself in $10$ years,is (Given $\log 2 = 0.6931$) (in $\%$)

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