1. The problem is to solve the exponential equations:\nc) Solve for $x$ in $4 \cdot 5^x = 210$.\nd) Solve for $x$ in $80 = 100 \cdot \left(\frac{1}{2}\right)^x$.\n\n2. For equation (c):\nDivide both sides by 4 to isolate $5^x$:\n$$5^x = \frac{210}{4} = 52.5$$\nTake the logarithm of both sides (using natural log $\ln$ or common log) to solve for $x$:\n$$\ln(5^x) = \ln(52.5)$$\nUsing the log power property:\n$$x \ln(5) = \ln(52.5)$$\nTherefore,\n$$x = \frac{\ln(52.5)}{\ln(5)}$$\nCalculate the values\n$$x \approx \frac{3.959}{1.609} \approx 2.46$$\n\n3. For equation (d):\nStart with:\n$$80 = 100 \cdot \left(\frac{1}{2}\right)^x$$\nDivide both sides by 100:\n$$0.8 = \left(\frac{1}{2}\right)^x$$\nTake the natural logarithm of both sides:\n$$\ln(0.8) = \ln\left(\left(\frac{1}{2}\right)^x\right)$$\nApply log power rule:\n$$\ln(0.8) = x \ln\left(\frac{1}{2}\right)$$\nSolve for $x$:\n$$x = \frac{\ln(0.8)}{\ln(\frac{1}{2})}$$\nNumerically,\n$$x \approx \frac{-0.223}{-0.693} \approx 0.322$$\n\nFinal answers:\nc) $x \approx 2.46$\nd) $x \approx 0.322$