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Let's solve the given equation step-by-step to find the approximate solutions. The equation is:
[tex]\[ \log_5(x + 5) = x^2 \][/tex]
Step 1: Understand the domain.
For the logarithmic function [tex]\(\log_5(x+5)\)[/tex] to be defined, the argument [tex]\(x + 5\)[/tex] must be positive:
[tex]\[ x + 5 > 0 \implies x > -5 \][/tex]
So, the solutions must come from the interval [tex]\(x > -5\)[/tex].
Step 2: Graphical or numerical approach to approximate the solutions.
Since the equation involves both a logarithmic function and a quadratic polynomial, an analytical approach might be complex. Instead, we will use a numerical and graphical approach to approximate the solutions.
Step 2.1: Plotting the functions.
a. Plot the logarithmic function [tex]\(y = \log_5(x + 5)\)[/tex].
b. Plot the quadratic function [tex]\(y = x^2\)[/tex].
We are looking for the points of intersection of these two curves.
Step 2.2: Check for intersections.
From the provided options, we'll evaluate and check:
Option 1: [tex]\(x \approx -0.93\)[/tex]
Calcualtion:
[tex]\[ \log_5(-0.93 + 5) = \log_5(4.07) \approx 0.83 \][/tex]
[tex]\[ (-0.93)^2 = 0.8649 \][/tex]
Since these values are close, [tex]\(\approx -0.93\)[/tex] is likely an approximate solution.
Option 2: [tex]\(x = 0\)[/tex]
Calculation:
[tex]\[ \log_5(0 + 5) = \log_5(5) = 1 \][/tex]
[tex]\[ 0^2 = 0 \][/tex]
Since this doesn't match, [tex]\(x = 0\)[/tex] is not a solution.
Option 3: [tex]\(x \approx 0.87\)[/tex]
Calculation:
[tex]\[ \log_5(0.87 + 5) = \log_5(5.87) \approx 1.23 \][/tex]
[tex]\[ (0.87)^2 = 0.7569 \][/tex]
These values don't match closely, so [tex]\(x \approx 0.87\)[/tex] isn't a solution.
Option 4: [tex]\(x \approx 1.06\)[/tex]
Calculation:
[tex]\[ \log_5(1.06 + 5) = \log_5(6.06) \approx 1.26 \][/tex]
[tex]\[ (1.06)^2 = 1.1236 \][/tex]
Since these values are close, [tex]\(\approx 1.06\)[/tex] is likely an approximate solution.
Conclusion:
Based on our evaluations, the approximate solutions to the equation [tex]\(\log_5(x + 5) = x^2\)[/tex] are:
- [tex]\( x \approx -0.93 \)[/tex]
- [tex]\( x \approx 1.06 \)[/tex]
Therefore, the answers which apply are:
- [tex]\( x \approx -0.93 \)[/tex]
- [tex]\( x \approx 1.06 \)[/tex]
[tex]\[ \log_5(x + 5) = x^2 \][/tex]
Step 1: Understand the domain.
For the logarithmic function [tex]\(\log_5(x+5)\)[/tex] to be defined, the argument [tex]\(x + 5\)[/tex] must be positive:
[tex]\[ x + 5 > 0 \implies x > -5 \][/tex]
So, the solutions must come from the interval [tex]\(x > -5\)[/tex].
Step 2: Graphical or numerical approach to approximate the solutions.
Since the equation involves both a logarithmic function and a quadratic polynomial, an analytical approach might be complex. Instead, we will use a numerical and graphical approach to approximate the solutions.
Step 2.1: Plotting the functions.
a. Plot the logarithmic function [tex]\(y = \log_5(x + 5)\)[/tex].
b. Plot the quadratic function [tex]\(y = x^2\)[/tex].
We are looking for the points of intersection of these two curves.
Step 2.2: Check for intersections.
From the provided options, we'll evaluate and check:
Option 1: [tex]\(x \approx -0.93\)[/tex]
Calcualtion:
[tex]\[ \log_5(-0.93 + 5) = \log_5(4.07) \approx 0.83 \][/tex]
[tex]\[ (-0.93)^2 = 0.8649 \][/tex]
Since these values are close, [tex]\(\approx -0.93\)[/tex] is likely an approximate solution.
Option 2: [tex]\(x = 0\)[/tex]
Calculation:
[tex]\[ \log_5(0 + 5) = \log_5(5) = 1 \][/tex]
[tex]\[ 0^2 = 0 \][/tex]
Since this doesn't match, [tex]\(x = 0\)[/tex] is not a solution.
Option 3: [tex]\(x \approx 0.87\)[/tex]
Calculation:
[tex]\[ \log_5(0.87 + 5) = \log_5(5.87) \approx 1.23 \][/tex]
[tex]\[ (0.87)^2 = 0.7569 \][/tex]
These values don't match closely, so [tex]\(x \approx 0.87\)[/tex] isn't a solution.
Option 4: [tex]\(x \approx 1.06\)[/tex]
Calculation:
[tex]\[ \log_5(1.06 + 5) = \log_5(6.06) \approx 1.26 \][/tex]
[tex]\[ (1.06)^2 = 1.1236 \][/tex]
Since these values are close, [tex]\(\approx 1.06\)[/tex] is likely an approximate solution.
Conclusion:
Based on our evaluations, the approximate solutions to the equation [tex]\(\log_5(x + 5) = x^2\)[/tex] are:
- [tex]\( x \approx -0.93 \)[/tex]
- [tex]\( x \approx 1.06 \)[/tex]
Therefore, the answers which apply are:
- [tex]\( x \approx -0.93 \)[/tex]
- [tex]\( x \approx 1.06 \)[/tex]
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