Solve the problem.
-The relationship between intensity I of light (in lumens) at a depth of $x$ feet in Lake $X$ is given by $\log \frac { \mathrm { I } } { 12 } = - 0.00264 x$. What is the intensity at a depth of 50 feet?(Round to the nearest ten-thousandth.)
A) $11.6408$ lumens
B) $0.0615$ lumens
C) $0.7379$ lumens
D) $8.8549$ lumens

Question

Quizplus · Accepted Answer

To find the intensity at a depth of 50 feet, substitute $x = 50$ into the equation: $\log \frac { \mathrm { I } } { 12 } = - 0.00264 \times 50$. This simplifies to $\log \frac { \mathrm { I } } { 12 } = -0.132$. Solving for $I$, we get $\frac { \mathrm { I } } { 12 } = 10^{-0.132}$, which gives $I = 12 \times 10^{-0.132} \approx 8.8549$ lumens.

Solve the Problem xxx Feet in Lake XXX Is Given By log⁡I12=−0.00264x\log \frac { \mathrm { I } } { 12 } = - 0.00264 xlog12I​=−0.00264x

Solve the Problem $x$ Feet in Lake $X$ Is Given By $\log \frac { \mathrm { I } } { 12 } = - 0.00264 x$