Solve the problem.
-Suppose $f ( x ) = 34.1 + 1.3 \log ( x + 1 )$ 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 36 .
A) $27.9 \mathrm {~m}$
B) $- 0.9 \mathrm {~m}$
C) $- 1.0 \mathrm {~m}$
D) $29.9 \mathrm {~m}$

Question

Quizplus · Accepted Answer

To find the depth where the salinity equals 36, set $f(x) = 36$ and solve for $x$:$$36 = 34.1 + 1.3 \log(x + 1)$$Subtract 34.1 from both sides:$$1.9 = 1.3 \log(x + 1)$$Divide by 1.3:$$\log(x + 1) = \frac{1.9}{1.3}$$Use the property of logarithms to solve for $x + 1$:$$x + 1 = 10^{\frac{1.9}{1.3}}$$Calculate the value:$$x + 1 \approx 10^{1.46153846}$$$$x + 1 \approx 29.0$$Subtract 1 to solve for $x$:$$x \approx 28.0$$Rounding to the nearest tenth gives $x \approx 27.9$ meters, which corresponds to choice A.

Solve the Problem f(x)=34.1+1.3log⁡(x+1)f ( x ) = 34.1 + 1.3 \log ( x + 1 )f(x)=34.1+1.3log(x+1)

Solve the Problem $f ( x ) = 34.1 + 1.3 \log ( x + 1 )$