Question
Determine whether each $x$ -value is a solution (or an approximate solution) of the equation.$4 e^{x-1}=60$(a) $x=1+\ln 15$(b) $x \approx 3.7081$(c) $x=\ln 16$
Step 1
We get $4e^{(1+\ln 15)-1}=60$, which simplifies to $4e^{\ln 15}=60$. Using the property of logarithms that $e^{\ln a}=a$, we get $4*15=60$. This is true, so $x=1+\ln 15$ is a solution. Show more…
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