IDNLearn.com: Where your questions meet expert answers and community support. Discover thorough and trustworthy answers from our community of knowledgeable professionals, tailored to meet your specific needs.

Solve for [tex]\( x \)[/tex].

[tex]\[ \log_5(x + 2) = 1 - \log_5(x + 6) \][/tex]

If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".

[tex]\[ x = \quad \square \][/tex]
[tex]\[ \square \][/tex] No solution


Sagot :

To solve the equation [tex]\(\log_5(x + 2) = 1 - \log_5(x + 6)\)[/tex], we'll proceed with the following steps:

1. Isolate the logarithms on one side:
[tex]\[ \log_5(x + 2) + \log_5(x + 6) = 1 \][/tex]

2. Use the properties of logarithms:
Recall the property of logarithms that states [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex]. Applying this property, we have:
[tex]\[ \log_5((x + 2)(x + 6)) = 1 \][/tex]

3. Convert the logarithmic equation to an exponential equation:
Recall that if [tex]\(\log_b(y) = c\)[/tex], then [tex]\(b^c = y\)[/tex]. So, we get:
[tex]\[ 5^1 = (x + 2)(x + 6) \][/tex]

4. Simplify and solve the quadratic equation:
[tex]\[ 5 = (x + 2)(x + 6) \][/tex]
Expand the right-hand side:
[tex]\[ 5 = x^2 + 6x + 2x + 12 \][/tex]
[tex]\[ 5 = x^2 + 8x + 12 \][/tex]
Subtract 5 from both sides to set the equation to zero:
[tex]\[ 0 = x^2 + 8x + 7 \][/tex]

5. Factor the quadratic equation:
We need to factor [tex]\(x^2 + 8x + 7\)[/tex]. Looking for two numbers that multiply to 7 and add up to 8, we get 1 and 7.
[tex]\[ 0 = (x + 1)(x + 7) \][/tex]

6. Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 1 = 0 \quad \text{or} \quad x + 7 = 0 \][/tex]
[tex]\[ x = -1 \quad \text{or} \quad x = -7 \][/tex]

7. Check for valid solutions:
* For [tex]\(x = -1\)[/tex]:
[tex]\[ \log_5(-1 + 2) = \log_5(1) = 0 \][/tex]
[tex]\[ 1 - \log_5(-1 + 6) = 1 - \log_5(5) = 1 - 1 = 0 \][/tex]
Both sides are equal, so [tex]\(x = -1\)[/tex] is a valid solution.

* For [tex]\(x = -7\)[/tex]:
[tex]\[ \log_5(-7 + 2) = \log_5(-5) \][/tex]
The logarithm of a negative number is not defined, so [tex]\(x = -7\)[/tex] is not a valid solution.

Thus, the only valid solution is:
[tex]\[ x = -1 \][/tex]
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.