Answer :
Let's solve each equation step-by-step and match it with the appropriate solution.
1. [tex]\[\ln(x+5) = \ln(x-1) + \ln(x+1)\][/tex]
Step-by-step solution:
- Use the properties of logarithms to combine the terms on the right side.
[tex]\[ \ln(x+5) = \ln((x-1)(x+1)) \][/tex]
- Since the natural logs are equal, the arguments must be equal.
[tex]\[ x + 5 = (x-1)(x+1) \][/tex]
- Expand and simplify the right side.
[tex]\[ x + 5 = x^2 - 1 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ x^2 - x - 6 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (x-3)(x+2) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
However, check for the domain (both must be greater than 1):
- [tex]\( x = 3 \)[/tex] is valid because [tex]\( x - 1 = 2 \)[/tex] and [tex]\( x + 1 = 4 \)[/tex].
- [tex]\( x = -2 \)[/tex] is not valid because [tex]\( x - 1 = -3 \)[/tex] (logarithm of negative number is not defined).
Solution: [tex]\( x = 3 \)[/tex]
2. [tex]\[\log_4(5x^2 + 2) = \log_4(x + 8)\][/tex]
Step-by-step solution:
- Since the bases of the logarithms are the same, we can equate the arguments.
[tex]\[ 5x^2 + 2 = x + 8 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ 5x^2 - x - 6 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (5x + 6)(x - 1) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = -\frac{6}{5} \quad \text{or} \quad x = 1 \][/tex]
However, check for the domain (both [tex]\(5x^2 + 2 \)[/tex] and [tex]\( x + 8 \)[/tex] must be positive):
- [tex]\( x = 1 \)[/tex] is valid.
- [tex]\( x = -\frac{6}{5} \)[/tex] is not valid because it makes the logarithm of a negative number.
Solution: [tex]\( x = 1 \)[/tex]
3. [tex]\[e^{x^2} = e^{4x + 5}\][/tex]
Step-by-step solution:
- Since the bases are the same, the exponents must be equal.
[tex]\[ x^2 = 4x + 5 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ x^2 - 4x - 5 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (x-5)(x+1) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = 5 \quad \text{or} \quad x = -1 \][/tex]
Both solutions are valid.
Solution: [tex]\( x = -1, 5 \)[/tex]
4. [tex]\[\log(x - 1) + \log(5x) = 2\][/tex]
Step-by-step solution:
- Use the properties of logarithms to combine the terms on the left side.
[tex]\[ \log((x-1) \cdot 5x) = 2 \][/tex]
- Simplify the argument of the logarithm.
[tex]\[ \log(5x^2 - 5x) = 2 \][/tex]
- Rewrite the logarithmic equation in exponential form.
[tex]\[ 5x^2 - 5x = 10^2 \][/tex]
[tex]\[ 5x^2 - 5x = 100 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ 5x^2 - 5x - 100 = 0 \][/tex]
[tex]\[ x^2 - x - 20 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (x-5)(x+4) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = 5 \quad \text{or} \quad x = -4 \][/tex]
However, check for the domain (both [tex]\( x-1 \)[/tex] and [tex]\(x\)[/tex] must be positive):
- [tex]\( x = 5 \)[/tex] is valid.
- [tex]\( x = -4 \)[/tex] is not valid because it makes the logarithm of a negative number.
Solution: [tex]\( x = 5 \)[/tex]
So, the matching of each equation with its solution is:
1. [tex]\(\ln(x + 5) = \ln(x - 1) + \ln(x + 1) \quad \rightarrow \quad x = 3\)[/tex]
2. [tex]\(\log_4(5x^2 + 2) = \log_4(x + 8) \quad \rightarrow \quad x = 1\)[/tex]
3. [tex]\(e^{x^2} = e^{4x + 5} \quad \rightarrow \quad x = -1, 5\)[/tex]
4. [tex]\(\log(x - 1) + \log(5x) = 2 \quad \rightarrow \quad x = 5\)[/tex]
1. [tex]\[\ln(x+5) = \ln(x-1) + \ln(x+1)\][/tex]
Step-by-step solution:
- Use the properties of logarithms to combine the terms on the right side.
[tex]\[ \ln(x+5) = \ln((x-1)(x+1)) \][/tex]
- Since the natural logs are equal, the arguments must be equal.
[tex]\[ x + 5 = (x-1)(x+1) \][/tex]
- Expand and simplify the right side.
[tex]\[ x + 5 = x^2 - 1 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ x^2 - x - 6 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (x-3)(x+2) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
However, check for the domain (both must be greater than 1):
- [tex]\( x = 3 \)[/tex] is valid because [tex]\( x - 1 = 2 \)[/tex] and [tex]\( x + 1 = 4 \)[/tex].
- [tex]\( x = -2 \)[/tex] is not valid because [tex]\( x - 1 = -3 \)[/tex] (logarithm of negative number is not defined).
Solution: [tex]\( x = 3 \)[/tex]
2. [tex]\[\log_4(5x^2 + 2) = \log_4(x + 8)\][/tex]
Step-by-step solution:
- Since the bases of the logarithms are the same, we can equate the arguments.
[tex]\[ 5x^2 + 2 = x + 8 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ 5x^2 - x - 6 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (5x + 6)(x - 1) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = -\frac{6}{5} \quad \text{or} \quad x = 1 \][/tex]
However, check for the domain (both [tex]\(5x^2 + 2 \)[/tex] and [tex]\( x + 8 \)[/tex] must be positive):
- [tex]\( x = 1 \)[/tex] is valid.
- [tex]\( x = -\frac{6}{5} \)[/tex] is not valid because it makes the logarithm of a negative number.
Solution: [tex]\( x = 1 \)[/tex]
3. [tex]\[e^{x^2} = e^{4x + 5}\][/tex]
Step-by-step solution:
- Since the bases are the same, the exponents must be equal.
[tex]\[ x^2 = 4x + 5 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ x^2 - 4x - 5 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (x-5)(x+1) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = 5 \quad \text{or} \quad x = -1 \][/tex]
Both solutions are valid.
Solution: [tex]\( x = -1, 5 \)[/tex]
4. [tex]\[\log(x - 1) + \log(5x) = 2\][/tex]
Step-by-step solution:
- Use the properties of logarithms to combine the terms on the left side.
[tex]\[ \log((x-1) \cdot 5x) = 2 \][/tex]
- Simplify the argument of the logarithm.
[tex]\[ \log(5x^2 - 5x) = 2 \][/tex]
- Rewrite the logarithmic equation in exponential form.
[tex]\[ 5x^2 - 5x = 10^2 \][/tex]
[tex]\[ 5x^2 - 5x = 100 \][/tex]
- Rearrange to form a quadratic equation.
[tex]\[ 5x^2 - 5x - 100 = 0 \][/tex]
[tex]\[ x^2 - x - 20 = 0 \][/tex]
- Factor the quadratic equation.
[tex]\[ (x-5)(x+4) = 0 \][/tex]
- Solve for [tex]\( x \)[/tex].
[tex]\[ x = 5 \quad \text{or} \quad x = -4 \][/tex]
However, check for the domain (both [tex]\( x-1 \)[/tex] and [tex]\(x\)[/tex] must be positive):
- [tex]\( x = 5 \)[/tex] is valid.
- [tex]\( x = -4 \)[/tex] is not valid because it makes the logarithm of a negative number.
Solution: [tex]\( x = 5 \)[/tex]
So, the matching of each equation with its solution is:
1. [tex]\(\ln(x + 5) = \ln(x - 1) + \ln(x + 1) \quad \rightarrow \quad x = 3\)[/tex]
2. [tex]\(\log_4(5x^2 + 2) = \log_4(x + 8) \quad \rightarrow \quad x = 1\)[/tex]
3. [tex]\(e^{x^2} = e^{4x + 5} \quad \rightarrow \quad x = -1, 5\)[/tex]
4. [tex]\(\log(x - 1) + \log(5x) = 2 \quad \rightarrow \quad x = 5\)[/tex]
