<Question>
Assume $0 < r < 3$.  Below are five equations for $x$.  Which equation has the largest solution $x$?

$\textbf{(A)}\ 3(1 + r)^x = 7\qquad \textbf{(B)}\ 3(1 + r/10)^x = 7\qquad \textbf{(C)}\ 3(1 + 2r)^x = 7$ $\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7$
</Question>

<think>
<Plan_1>
Okay, let's see. The problem is asking which of the five given equations has the largest solution for x, given that 0 < r < 3. All the equations are of the form 3 times something raised to the power x equals 7. So, they're all exponential equations. I need to figure out, for each equation, what x would be, and then determine which one is the largest.

First, maybe I should recall how to solve exponential equations. The general form is a*b^x = c, right? So, to solve for x, we can use logarithms. Let me write down the steps.

<prompt_1.1> Recall the method to solve exponential equations and prepare to apply it to each given equation. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
For an equation like 3*(1 + something)^x = 7, we can divide both sides by 3 to get (1 + something)^x = 7/3. Then take the natural logarithm of both sides: ln((1 + something)^x) = ln(7/3). Using the logarithm power rule, that becomes x*ln(1 + something) = ln(7/3). Then, solving for x gives x = ln(7/3)/ln(1 + something).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, the solution x is the natural logarithm of 7/3 divided by the natural logarithm of (1 + the term involving r). Since all the equations have the same structure, the key factor here is how big or small the base (1 + something) is. Because the denominator in the x expression is ln(1 + something), if the base is larger, the denominator is larger, which would make x smaller. Conversely, if the base is smaller, the denominator is smaller, making x larger. But wait, this is assuming that the base is greater than 1.

Wait, since 0 < r < 3, let's check each equation's base:

Looking at the equations:

A: 1 + r

<prompt_2.1> Analyze the bases of the equations and determine how they affect the solution x. </prompt_2.1>
-----

But wait, but in equation C, the base is 1 + 2r. Since r is between 0 and 3, 2r is between 0 and 6, so 1 + 2r is between 1 and 7. So, for r approaching 3, 1 + 2r approaches 7. So, the base is larger than in A, B, D, E. Therefore, equation C's base is the largest, which would mean that x is the smallest, since the denominator is the largest. Therefore, C has the smallest x.

So, so far, the order from largest base to smallest base is: C > A > D > B > E. Therefore, the order of x from smallest to largest would be E < B < D < A < C. So, the largest x is in equation E. But wait, that contradicts my previous thought. Wait, maybe I confused the order. Let me think again.

<prompt_2.2> Re-evaluate the order of bases and their impact on x, and confirm the largest x. </prompt_2.2>
-----

But wait, let me check with specific numbers. Let's pick a value of r between 0 and 3, say r = 1.

<prompt_2.3> Test specific values of r to confirm which equation has the largest x. </prompt_2.3>

Another example, take r approaching 0. Let's say r = 0.1.

<prompt_2.4> Test another value of r to confirm which equation has the largest x. </prompt_2.4>

Wait, but let's check another value, say r = 2.

<prompt_2.5> Test additional values of r to confirm which equation has the largest x. </prompt_2.5>

But let's check with r approaching 3. Let's take r = 2.9.

<prompt_2.6> Test values of r approaching the upper limit to confirm which equation has the largest x. </prompt_2.6>

Alternatively, maybe we can analyze the derivative or something else? Wait, but since all functions are exponential, and since the base is the critical factor here. Since 3*b^x =7, solving for x gives x = ln(7/3)/ln(b). So, if b increases, ln(b) increases, so x decreases. Therefore, the larger the base b, the smaller the x. Therefore, to maximize x, we need the smallest possible base b. Therefore, the equation with the smallest base will have the largest x. Therefore, instead of comparing the bases to find the largest x, we can find the equation with the smallest base. Therefore, perhaps this is a simpler way: find which equation has the smallest base, and that will have the largest x.

So, bases:

A: 1 + r

<prompt_2.7> Compare the bases of the equations to determine which has the smallest base. </prompt_2.7>
</Plan_2>

<Execution_2>
<execution_2.1>
B: 1 + r/10

C: 1 + 2r

D: 1 + sqrt(r)

E: 1 + 1/r

So, depending on r, these bases can vary. Since r is between 0 and 3, let's analyze each base:

For equation A: 1 + r. Since r is between 0 and 3, the base here is between 1 and 4.

Equation B: 1 + r/10. Since r is up to 3, r/10 is up to 0.3, so the base is between 1 and 1.3.

Equation C: 1 + 2r. Since r is up to 3, 2r is up to 6, so the base is between 1 and 7.

Equation D: 1 + sqrt(r). Since r is up to 3, sqrt(r) is up to sqrt(3) ≈ 1.732, so the base is between 1 and approximately 2.732.

Equation E: 1 + 1/r. Since r is between 0 and 3, 1/r is greater than 1/3, so the base is greater than 1 + 1/3 ≈ 1.333. But as r approaches 0, 1/r approaches infinity, so the base can be very large.
</execution_2.1>

<execution_2.2>
Wait, actually, higher base would lead to a smaller x, as x = ln(7/3)/ln(base). So, if the base is larger, the denominator is larger, so x is smaller. Therefore, the equation with the smallest base will have the largest x. Therefore, to find the largest x, we need the equation with the smallest base (1 + something). Therefore, among the bases, the smallest base will give the largest x. Therefore, the problem reduces to finding which equation has the smallest base (1 + something), because that will give the largest x.
</execution_2.2>

<execution_2.3>
For r = 1:

A: 1 + 1 = 2

B: 1 + 1/10 = 1.1

C: 1 + 2*1 = 3

D: 1 + sqrt(1) = 2

E: 1 + 1/1 = 2

So, bases are 2, 1.1, 3, 2, 2. So, the largest base is C (3), then A, D, E (both 2), and B (1.1). So, in this case, the x values would be:

x_A = ln(7/3)/ln(2) ≈ ln(2.333)/0.693 ≈ 0.847/0.693 ≈ 1.22

x_B = ln(7/3)/ln(1.1) ≈ 0.847/0.0953 ≈ 8.89

x_C = ln(7/3)/ln(3) ≈ 0.847/1.0986 ≈ 0.771

x_D = same as A, since base is 2: ≈1.22

x_E = same as A and D: ≈1.22

So, in this case, equation B has the largest x, which is 8.89, while C has the smallest x, 0.771. So, for r=1, the answer is B.
</execution_2.3>

<execution_2.4>
For r = 1:

Equation A: 1 + 1 = 2

Equation B: 1 + 1/10 = 1.1

Equation C: 1 + 2*1 = 3

Equation D: 1 + sqrt(1) = 2

Equation E: 1 + 1/1 = 2

So, bases here are 2, 1.1, 3, 2, 2. So, the largest base is 3 (equation C), then 2s (D and E), then 1.1 (B), then 2 (A). Wait, but in this case, equation C has the largest base, so x would be the smallest. Equations D and E have the same base, so their x would be the same? Wait, but in this case, 1 + something is the same for D and E. For D, sqrt(1) is 1, so 1 + 1 = 2. For E, 1/r is 1, so 1 + 1 = 2. So, yes, same base. Therefore, same x. So, in this case, x for A is ln(7/3)/ln(2), x for B is ln(7/3)/ln(1.1), which is larger because ln(1.1) is smaller (since 1.1 < 2), so ln(1.1) is smaller positive number, so x is larger. Similarly, x for C is ln(7/3)/ln(3), which is smaller because ln(3) is larger. So, in this case, x order is B > A > D=E > C.
</execution_2.4>

<execution_2.5>
Let's compute x for each equation when r = 1.

For A: 3*(1 + 1)^x = 7 => 3*2^x =7 => 2^x = 7/3 ≈2.333. x = ln(7/3)/ln(2) ≈ ln(2.333)/0.693 ≈0.847/0.693≈1.22.

For B: 3*(1 + 1/10)^x =7 => 3*(1.1)^x=7 =>1.1^x≈2.333. x= ln(2.333)/ln(1.1)≈0.847/0.0953≈8.89.

For C:3*(1 +2*1)^x=7 =>3*3^x=7 =>3^x≈2.333. x= ln(2.333)/ln(3)≈0.847/1.098≈0.771.

For D:3*(1 + sqrt(1))^x=7 =>3*(2)^x=7. Wait, sqrt(1)=1, so same as A. So, x≈1.22.

For E:3*(1 +1/1)^x=7 =>3*(2)^x=7. Again, same base as A and D, so x≈1.22.

Wait, but when r=1, equations A, D, E all have base 2, so same x≈1.22. Equation B has base 1.1, which gives x≈8.89, which is much larger. Equation C has base 3, x≈0.771. So, in this case, equation B has the largest x.
</execution_2.5>

<execution_2.6>
Then:

Equation A: 1 + r = 4. So, 3*4^x =7. x= ln(7/3)/ln(4) ≈ ln(2.333)/1.386 ≈ 0.847/1.386 ≈ 0.613.

Equation B: 1 + r/10 = 1 + 2.9/10 = 1.29. So, 3*(1.29)^x =7. x= ln(7/3)/ln(1.29) ≈ 0.847/0.247 ≈ 3.43.

Equation C: 1 + 2r = 1 + 5.8 =6.8. 3*(6.8)^x=7. x= ln(7/3)/ln(6.8) ≈0.847/1.916≈0.442.

Equation D:1 + sqrt(2.9)≈1 +1.702≈2.702. 3*(2.702)^x=7. x= ln(7/3)/ln(2.702)≈0.847/1.000≈0.847.

Equation E:1 +1/2.9≈1 +0.344≈1.344. 3*(1.344)^x=7. x= ln(7/3)/ln(1.344)≈0.847/0.297≈2.857.

So, with r=2.9, the x values are approximately:

A:0.613, B:3.43, C:0.442, D:0.847, E:2.857.

So, the largest x here is B. So, in this case, the order is B > E > D > A > C. So, B is the largest. But when r was 1, B was 1.344, D was 1.25, E was 1.667. So, E was the largest. So, which one is larger depends on r.
</execution_2.6>

<execution_2.7>
Since 0 < r <3, so base A is between 1 and 4.

B: 1 + r/10. Since r is up to 3, r/10 is up to 0.3, so base B is between 1 and 1.3.

C: 1 + 2r. Since r is up to 3, 2r is up to 6, so base C is between 1 and 7.

D: 1 + sqrt(r). Since r is up to 3, sqrt(r) is up to sqrt(3) ≈ 1.732, so base D is between 1 and ≈2.732.

E: 1 + 1/r. Since r is between 0 and 3, 1/r is between 1/3 and infinity. But since r is less than 3, 1/r is greater than 1/3. However, since r is positive, 1/r is positive. So, base E is 1 + something greater than 1/3, so base E is greater than 4/3. But when r approaches 0, 1/r approaches infinity, so base E approaches infinity. However, r is strictly greater than 0, so base E is greater than 1. But depending on r, it can be larger or smaller.

Wait, but for 0 < r <3, 1/r is greater than 1/3, so base E is 1 + 1/r, which is between 4/3 ≈1.333 and infinity. Wait, but when r approaches 3, base E is 1 + 1/3 ≈1.333. When r approaches 0, base E approaches infinity. So, base E can be as large as possible. However, in the given problem, r is fixed between 0 and 3, but we are to consider which equation has the largest solution x for a given r. Wait, but the problem says "Assume 0 < r <3. Below are five equations for x. Which equation has the largest solution x"? So, for a given r in (0,3), which equation's x is the largest. So, since x is inversely related to ln(b), where b is the base, then the equation with the smallest base will have the largest x. Therefore, for a given r, we need to find which base is the smallest among the five equations. Therefore, the equation with the smallest base will yield the largest x. Therefore, we need to compare the bases for each equation given a specific r, but since r is arbitrary in (0,3), we need to see which of the bases is the smallest across all possible r in (0,3). Wait, but actually, for a given r, we can compute each base and see which is the smallest. However, since r is fixed but unknown, the question is which equation's base is the smallest for any r in (0,3). Wait, but maybe not. Wait, the problem states "Assume 0 < r <3. Below are five equations for x. Which equation has the largest solution x"? So, for a given r in (0,3), which equation's x is the largest. So, for each r in (0,3), compute the five bases, find which is the smallest, and that equation's x is the largest. Therefore, the answer is the equation with the smallest base. Therefore, we need to find, among the five equations, which one has the smallest base for any given r in (0,3). So, let's compare the bases:

Equation A: 1 + r

Equation B: 1 + r/10

Equation C: 1 + 2r

Equation D: 1 + sqrt(r)

Equation E: 1 + 1/r

So, for each r in (0,3), which of these is the smallest?
</execution_2.7>
</Execution_2>

<Plan_3>
Based on execution_2.7:
Let's analyze:

First, note that for 0 < r <3:

- A: 1 + r. Since r is between 0 and 3, A is between 1 and 4.

- B: 1 + r/10. Since r/10 is between 0 and 0.3, B is between 1 and 1.3.

- C: 1 + 2r. Since 2r is between 0 and 6, C is between 1 and 7.

- D: 1 + sqrt(r). Since sqrt(r) is between 0 and sqrt(3) ≈1.732, D is between 1 and ≈2.732.

- E: 1 + 1/r. Since r is between 0 and 3, 1/r is between 1/3 and infinity. So, E is between 4/3 ≈1.333 and infinity.

So, comparing all five:

For a given r, we need to see which of A, B, C, D, E is the smallest.
Let's compare B with others.

<prompt_3.1> Compare the bases of the equations to determine which is the smallest. </prompt_3.1>
-----

But let's check when r is approaching 3. For example, take r = 2.5.

<prompt_3.2> Re-evaluate the comparison of bases for different values of r. </prompt_3.2>
-----

Therefore, the answer is B.

But let's check with another value of r, say r = 0.5.

<prompt_3.3> Confirm the conclusion by testing another value of r. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
Since B is 1 + r/10, and D is 1 + sqrt(r). Let's compare r/10 and sqrt(r). For 0 < r <3:

sqrt(r) vs r/10.

Let’s solve sqrt(r) = r/10. Let’s set sqrt(r) = r/10.

Let’s square both sides: r = r² / 100 => r² -100r =0 => r(r -100)=0. So, solutions at r=0 and r=100. Since we have 0 < r <3, sqrt(r) > r/10 because for 0 < r <100, sqrt(r) > r/10. For example, take r=1: sqrt(1)=1, 1/10=0.1, so sqrt(r) > r/10. Similarly, at r=0.5: sqrt(0.5)≈0.707, 0.5/10=0.05. So, sqrt(r) > r/10. Therefore, in 0 < r <3, sqrt(r) > r/10. Therefore, 1 + sqrt(r) > 1 + r/10. Therefore, D > B.

Similarly, compare B with E: B is 1 + r/10, E is 1 + 1/r. So, which is smaller: r/10 vs 1/r?

Set r/10 = 1/r => r² =10 => r= sqrt(10)≈3.16. But since r <3, in 0 < r <3, r/10 <1/r. Because for r < sqrt(10), r/10 <1/r. For example, take r=2: 2/10=0.2, 1/2=0.5. So, r/10 <1/r. Therefore, in 0 < r <3, r/10 <1/r. Therefore, 1 + r/10 <1 +1/r. Therefore, B < E.

Similarly, compare B with A: B is 1 + r/10, A is 1 + r. Since r/10 < r (for r >0), so B < A.

Compare B with C: B is 1 + r/10, C is 1 +2r. For 0 < r <3, r/10 <2r (since r/10 <2r => 1/10 <2 => true for r >0). Therefore, B < C.

Therefore, for all r in (0,3), B is smaller than A, C, D, E. Wait, is that possible?

Wait, but let's check when r approaches 0. For very small r, say r=0.1:



Wait, but when r approaches 3, let's check:

r=2.5:
</execution_3.1>

<execution_3.2>
Then:

A: 1 + 2.5 = 3.5

B: 1 + 0.25 = 1.25

C: 1 + 5 = 6

D: 1 + sqrt(2.5) ≈1 + 1.581 ≈2.581

E: 1 + 2 = 3

So, the bases are 3.5, 1.25, 6, 2.581, 3. So, the smallest is B (1.25). So, x for B would be ln(7/3)/ln(1.25) ≈0.847/0.223≈3.79. For A: ln(7/3)/ln(3.5)≈0.847/1.255≈0.674. For C: ln(7/3)/ln(6)≈0.847/1.792≈0.473. For D: ln(7/3)/ln(2.581)≈0.847/0.951≈0.891. For E: ln(7/3)/ln(3)≈0.847/1.098≈0.771. So, again, B has the largest x.
</execution_3.2>

<execution_3.3>
For r = 0.5:

Equation A: 1 + 0.5 = 1.5

Equation B: 1 + 0.05 = 1.05

Equation C: 1 + 1.0 = 2.0

Equation D: 1 + sqrt(0.5) ≈1 + 0.707 ≈1.707

Equation E: 1 + 2 = 2.0 (since 1/r = 2)

So, bases: 1.5, 1.05, 2.0, 1.707, 2.0. The smallest is 1.05 (B). So, x would be largest for B again.

Another example: r approaching 0. Let's take r = 0.1.

Equation A: 1.1

Equation B: 1 + 0.01 = 1.01

Equation C: 1 + 0.2 = 1.2

Equation D: 1 + sqrt(0.1) ≈1 + 0.316 ≈1.316

Equation E: 1 + 10 = 11

So, bases: 1.1, 1.01, 1.2, 1.316, 11. The smallest is 1.01 (B). So, x is largest for B.

Another case: Let's take r = 2. Let's see.

Equation A: 1 + 2 = 3

Equation B: 1 + 0.2 = 1.2

Equation C: 1 + 4 = 5

Equation D: 1 + sqrt(2) ≈2.414

Equation E: 1 + 0.5 = 1.5

So, bases: 3, 1.2, 5, 2.414, 1.5. The smallest is 1.2 (B). So, x is largest for B.

Wait, but when r = 2.9, as in the previous example, the smallest base was B (1.29), D (≈2.702), E (≈1.344), A (≈2.702), C (≈6.8). So, B was still the smallest. So, in all these cases, B has the smallest base, leading to the largest x. Therefore, regardless of the value of r in (0,3), equation B always has the smallest base, hence the largest x. Therefore, the answer is B.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.3:
Wait, but let me check when r is very close to 3, like r = 2.99.

<prompt_4.1> Perform a final verification to ensure equation B is consistently the largest. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Then:

A: 1 + 2.99 = 3.99

B: 1 + 2.99/10 = 1.299

C: 1 + 2*2.99 = 6.99

D: 1 + sqrt(2.99) ≈1 + 1.729 ≈2.729

E: 1 + 1/2.99 ≈1 + 0.334 ≈1.334

So, the bases are 3.99, 1.299, 6.99, 2.729, 1.334. The smallest base is 1.299 (B). So, x is largest here. So, even when r is approaching 3, B still has the smallest base. Therefore, x is largest for equation B.

Therefore, regardless of the value of r in (0,3), equation B (3(1 + r/10)^x =7) has the smallest base, leading to the largest solution x. Therefore, the answer is B.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
**Final Answer**
\boxed{B}
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Thus, the equation with the largest solution for $x$ is $\boxed{B}$.
</Final_answer>
</think>