Solve the problem.
-Suppose $f ( x ) = 28.4 + 1.3 \log ( x + 1 )$ models salinity of ocean water to depths of 1000 meters at a certain latitude. $x$ is the depth in meters and $f ( x )$ is in grams of salt per kilogram of seawater. Approximate the salinity (to the nearest hundredth) when the depth is 383 meters.
A) $74.70 \mathrm {~g} / \mathrm { kg }$
B) $31.76 \mathrm {~g} / \mathrm { kg }$
C) $72.10 \mathrm {~g} / \mathrm { kg }$
D) $- 25.04 \mathrm {~g} / \mathrm { kg }$

Question

Quizplus · Accepted Answer

To find the salinity at 383 meters, substitute $x = 383$ into the given model: $f(383) = 28.4 + 1.3 \log(383 + 1)$. Simplifying, $f(383) = 28.4 + 1.3 \log(384)$. Using a calculator, this gives approximately $31.76 \mathrm {~g} / \mathrm { kg }$.

Solve the Problem f(x)=28.4+1.3log⁡(x+1)f ( x ) = 28.4 + 1.3 \log ( x + 1 )f(x)=28.4+1.3log(x+1)

Solve the Problem $f ( x ) = 28.4 + 1.3 \log ( x + 1 )$