Question

hw unit 15 - logarithmic and exponential equations; applications
question 33 of 35 (1 point) | question attempt: 1 of unlimited
subtract 2 from both sides to isolate the exponential expression.
$e^{0.02k} = 57$
the exponential expression has a base of $e$, so it is convenient to take th
$ln e^{0.02k} = ln 57$
use the power property of logarithms.
$0.02k ln e = ln 57$
part: 1 / 2
part 2 of 2
simplify (recall that $ln e = 1$).
$\square = ln 57$

Explanation:

Step1: Recall the property of ln(e)

We know that $\ln e = 1$. From the previous step, we have $0.02k \ln e=\ln 57$.

Step2: Substitute ln(e) with 1

Substitute $\ln e$ with 1 in the equation $0.02k \ln e=\ln 57$. So we get $0.02k\times1=\ln 57$, which simplifies to $0.02k = \ln 57$.

Answer:

$0.02k$