Subjects algebra

Exponential Equation Cd523D

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Exponential Equation Cd523D


1. **State the problem:** Solve the equation $$2000(0.8)^{\frac{x}{2}} = 1000(1.15)^x$$ for $x$. 2. **Rewrite the equation:** Divide both sides by 1000 to simplify: $$2(0.8)^{\frac{x}{2}} = (1.15)^x$$ 3. **Express powers with a common base or use logarithms:** Take the natural logarithm (ln) of both sides: $$\ln\left(2(0.8)^{\frac{x}{2}}\right) = \ln\left((1.15)^x\right)$$ 4. **Use logarithm properties:** $$\ln 2 + \ln\left((0.8)^{\frac{x}{2}}\right) = x \ln(1.15)$$ $$\ln 2 + \frac{x}{2} \ln 0.8 = x \ln 1.15$$ 5. **Group terms with $x$ on one side:** $$\ln 2 = x \ln 1.15 - \frac{x}{2} \ln 0.8$$ $$\ln 2 = x \left(\ln 1.15 - \frac{1}{2} \ln 0.8\right)$$ 6. **Solve for $x$:** $$x = \frac{\ln 2}{\ln 1.15 - \frac{1}{2} \ln 0.8}$$ 7. **Calculate the values:** $$\ln 2 \approx 0.6931$$ $$\ln 1.15 \approx 0.1398$$ $$\ln 0.8 \approx -0.2231$$ Substitute: $$x = \frac{0.6931}{0.1398 - \frac{1}{2}(-0.2231)} = \frac{0.6931}{0.1398 + 0.11155} = \frac{0.6931}{0.25135} \approx 2.757$$ **Final answer:** $$x \approx 2.76$$