The intensity of light at distance from a source is modeled by where , , and . What is in terms of , , and ?

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SAT Nonlinear Equations & Systems Question 232 | Free SAT Prep | Aniko

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Question 232·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The intensity $I$ of light at distance $d$ from a source is modeled by

I = \frac{I_0}{(1+kd)^2}

where $I_0>0$ , $k>0$ , and $0<I<I_0$ . What is $d$ in terms of $I$ , $I_0$ , and $k$ ?

For equations where the variable you want is inside a squared term in the denominator, treat it systematically: (1) multiply both sides to clear the denominator, (2) divide to isolate the squared expression, (3) take the square root of both sides—being careful about the $\pm$ and using any given domain or context to pick the correct sign—and then (4) solve the resulting simple linear equation. On the SAT, watch for common traps: flipping a fraction when you divide, forgetting to take a square root, or reversing the order in a subtraction; quickly checking which choices include a necessary square root or would make a distance negative can help you eliminate wrong answers fast.

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Get rid of the denominator

Your first goal is to remove the fraction. What can you multiply both sides of $I = \dfrac{I_0}{(1+kd)^2}$ by to eliminate the denominator $(1+kd)^2$ ?

Isolate the squared expression

After clearing the denominator, rearrange the equation so that $(1+kd)^2$ is by itself on one side. What should be on the other side of the equation?

Undo the square carefully

Once you have $(1+kd)^2 =$ (something), take the square root of both sides. Remember this gives a $\pm$ result, but think about the given conditions $I_0>0$ , $k>0$ , $0<I<I_0$ , and the fact that $d$ is a distance to decide which sign makes sense. Then solve the resulting linear equation for $d$ .

Desmos Guide

Set up parameters for the model

In Desmos, create sliders for I0, k, and I. Choose positive values with 0 < I < I0 (for example, set I0 = 10, k = 2, and I = 4). These will act as specific test values for the parameters in the formula.

Enter each option as a possible expression for d

For each answer choice, type a separate expression in Desmos, such as dA = (1/k)*(sqrt(I0/I) - 1), dB = (1/k)*(1 - sqrt(I/I0)), etc., matching the right-hand side of each option. These give numerical values for d based on your chosen I0, k, and I.

Check which expression satisfies the original equation

For each candidate (e.g., dA, dB, ...), define IcheckA = I0/(1 + k*dA)^2, IcheckB = I0/(1 + k*dB)^2, and so on. Compare each Icheck value with your slider I. The correct formula for $d$ will produce Icheck exactly equal to I (or matching for multiple different slider settings), while the incorrect ones will not.

Step-by-step Explanation

Understand the goal and the structure of the equation

We are given

I = \frac{I_0}{(1+kd)^2}

and asked to solve this equation for $d$ in terms of $I$ , $I_0$ , and $k$ . Notice that $d$ is inside the squared term in the denominator, so we will first need to remove the fraction and the square before isolating $d$ .

Clear the denominator and isolate the squared term

Multiply both sides by $(1+kd)^2$ to eliminate the denominator:

I(1+kd)^2 = I_0.

Now divide both sides by $I$ (which is positive, so this is allowed):

(1+kd)^2 = \frac{I_0}{I}.

Now the squared expression involving $d$ is isolated.

Undo the square with a square root and choose the correct sign

Take the square root of both sides to undo the square:

1+kd = \pm \sqrt{\frac{I_0}{I}}.

We must decide between the $+$ and $-$ signs. We are told $I_0>0$ , $k>0$ , $0<I<I_0$ , and $d$ represents a distance (so $d\ge 0$ ):

Because $0<I<I_0$ , the fraction $\dfrac{I_0}{I}$ is greater than $1$ , so $\sqrt{\dfrac{I_0}{I}} > 1$ .
Since $k>0$ and $d\ge0$ , we have $kd\ge0$ , so $1+kd \ge 1 > 0$ .

Therefore $1+kd$ must be positive, so we choose the positive root:

1+kd = \sqrt{\frac{I_0}{I}}.

Solve the resulting linear equation for d (final answer)

Now solve $1+kd = \sqrt{\dfrac{I_0}{I}}$ for $d$ .

Subtract $1$ from both sides:

kd = \sqrt{\frac{I_0}{I}} - 1.

Then divide both sides by $k$ :

d = \frac{1}{k}\left(\sqrt{\frac{I_0}{I}} - 1\right).

This matches answer choice A.

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Step-by-step Explanation

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Step-by-step Explanation