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 depth (to the nearest tenth of a meter) where the salinity equals 29 .
A) $1.9 \mathrm {~m}$
B) $- 0.9 \mathrm {~m}$
C) $- 0.7 \mathrm {~m}$
D) $3.9 \mathrm {~m}$

Question

Quizplus · Accepted Answer

To find the depth where the salinity equals 29, set $f(x) = 29$ and solve for $x$: $29 = 28.4 + 1.3 \log(x + 1)$. Subtract 28.4 from both sides to get $0.6 = 1.3 \log(x + 1)$, then divide by 1.3 to get $\log(x + 1) = \frac{0.6}{1.3}$. Solve for $x + 1$ by taking the antilog, leading to $x + 1 = 10^{\frac{0.6}{1.3}}$, and finally solve for $x$ to find the depth. The calculation gives a depth close to 1.9 meters.

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