Question

Consider the equation 7 • e^0.1t = 47.Solve the equation for t. Express the solution as a logarithm in base-e.t = ___Approximate the value of t . Round your answer to the nearest thousandth.t ≈ ___

Answer 1

Answer:$\begin{gathered} t=10\log_e47 \\ \\ \approx38.501 \end{gathered}$

Explanation:

Given the equation:

$e^{0.1t}=47$

Applying logarithm to both sides:

$\begin{gathered} \log_ee^{0.1t}=\log_e47 \\ \\ 0.1t\log_ee=\log_e47 \\ \\ 0.1t=\log_e47 \\ \\ t=\frac{1}{0.1}\log_e47 \\ \\ t=10\log_e47 \end{gathered}$

Aproximately, we have:

$t=38.501$