The number of bacteria \( P(h) \) in a certain population increases according to the following function, where time \( h \) is measured in hours. \[ P(h)=2600 e^{0.06 h} \] How many hours will it take for the number of bacteria to reach 3200 ? Round your answer to the nearest tenth, and do not round any intermediate computations.

We begin with the equation for the number of bacteria: \[ 2600 e^{0.06h} = 3200. \] To solve for \( h \), first divide both sides by \( 2600 \): \[ e^{0.06h} = \frac{3200}{2600} = \frac{32}{26} = \frac{16}{13}. \] Next, take the natural logarithm of both sides: \[ \ln\left(e^{0.06h}\right) = \ln\left(\frac{16}{13}\right). \] Using the logarithmic identity \( \ln(e^x) = x \), we have: \[ 0.06h = \ln\left(\frac{16}{13}\right). \] Now, solve for \( h \): \[ h = \frac{\ln\left(\frac{16}{13}\right)}{0.06}. \] Using a calculator to evaluate the logarithm: \[ \ln\left(\frac{16}{13}\right) \approx 0.207014. \] Then, \[ h \approx \frac{0.207014}{0.06} \approx 3.450233. \] Rounding to the nearest tenth: \[ h \approx 3.5 \text{ hours}. \]