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To solve the equation:
[tex]\[ \log_6(x + 11) + \log_6(x + 6) = 2 \][/tex]
we'll follow these steps:
1. Combine the logarithms using the properties of logarithms:
The sum of two logarithms with the same base can be combined into a single logarithm:
[tex]\[ \log_6(x + 11) + \log_6(x + 6) = \log_6\left((x + 11)(x + 6)\right) \][/tex]
So the equation becomes:
[tex]\[ \log_6((x + 11)(x + 6)) = 2 \][/tex]
2. Use the definition of logarithms to eliminate the logarithm:
Recall that if [tex]\(\log_b(y) = c\)[/tex], then [tex]\(y = b^c\)[/tex]. Applying this to our equation,
[tex]\[ (x + 11)(x + 6) = 6^2 \][/tex]
Simplifying further,
[tex]\[ (x + 11)(x + 6) = 36 \][/tex]
3. Expand and simplify the equation:
Expand the left-hand side of the equation:
[tex]\[ x^2 + 6x + 11x + 66 = 36 \][/tex]
Combine like terms:
[tex]\[ x^2 + 17x + 66 = 36 \][/tex]
4. Bring all terms to one side to form a quadratic equation:
[tex]\[ x^2 + 17x + 30 = 0 \][/tex]
5. Solve the quadratic equation:
To solve [tex]\(x^2 + 17x + 30 = 0\)[/tex], we can use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 17\)[/tex], and [tex]\(c = 30\)[/tex]. Substituting these values in, we get:
[tex]\[ x = \frac{-17 \pm \sqrt{17^2 - 4 \cdot 1 \cdot 30}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{-17 \pm \sqrt{289 - 120}}{2} \][/tex]
[tex]\[ x = \frac{-17 \pm \sqrt{169}}{2} \][/tex]
[tex]\[ x = \frac{-17 \pm 13}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-17 + 13}{2} = \frac{-4}{2} = -2 \][/tex]
and
[tex]\[ x = \frac{-17 - 13}{2} = \frac{-30}{2} = -15 \][/tex]
6. Check for the validity of the solutions in the original equation:
For [tex]\(x = -2\)[/tex]:
[tex]\[ \log_6(-2 + 11) + \log_6(-2 + 6) = \log_6(9) + \log_6(4) \][/tex]
This is valid as both arguments of the logarithms are positive.
For [tex]\(x = -15\)[/tex]:
[tex]\[ \log_6(-15 + 11) + \log_6(-15 + 6) = \log_6(-4) + \log_6(-9) \][/tex]
This is invalid since the arguments of the logarithms are negative.
Therefore, the only valid solution is:
[tex]\[ x = -2 \][/tex]
[tex]\[ \log_6(x + 11) + \log_6(x + 6) = 2 \][/tex]
we'll follow these steps:
1. Combine the logarithms using the properties of logarithms:
The sum of two logarithms with the same base can be combined into a single logarithm:
[tex]\[ \log_6(x + 11) + \log_6(x + 6) = \log_6\left((x + 11)(x + 6)\right) \][/tex]
So the equation becomes:
[tex]\[ \log_6((x + 11)(x + 6)) = 2 \][/tex]
2. Use the definition of logarithms to eliminate the logarithm:
Recall that if [tex]\(\log_b(y) = c\)[/tex], then [tex]\(y = b^c\)[/tex]. Applying this to our equation,
[tex]\[ (x + 11)(x + 6) = 6^2 \][/tex]
Simplifying further,
[tex]\[ (x + 11)(x + 6) = 36 \][/tex]
3. Expand and simplify the equation:
Expand the left-hand side of the equation:
[tex]\[ x^2 + 6x + 11x + 66 = 36 \][/tex]
Combine like terms:
[tex]\[ x^2 + 17x + 66 = 36 \][/tex]
4. Bring all terms to one side to form a quadratic equation:
[tex]\[ x^2 + 17x + 30 = 0 \][/tex]
5. Solve the quadratic equation:
To solve [tex]\(x^2 + 17x + 30 = 0\)[/tex], we can use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 17\)[/tex], and [tex]\(c = 30\)[/tex]. Substituting these values in, we get:
[tex]\[ x = \frac{-17 \pm \sqrt{17^2 - 4 \cdot 1 \cdot 30}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{-17 \pm \sqrt{289 - 120}}{2} \][/tex]
[tex]\[ x = \frac{-17 \pm \sqrt{169}}{2} \][/tex]
[tex]\[ x = \frac{-17 \pm 13}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-17 + 13}{2} = \frac{-4}{2} = -2 \][/tex]
and
[tex]\[ x = \frac{-17 - 13}{2} = \frac{-30}{2} = -15 \][/tex]
6. Check for the validity of the solutions in the original equation:
For [tex]\(x = -2\)[/tex]:
[tex]\[ \log_6(-2 + 11) + \log_6(-2 + 6) = \log_6(9) + \log_6(4) \][/tex]
This is valid as both arguments of the logarithms are positive.
For [tex]\(x = -15\)[/tex]:
[tex]\[ \log_6(-15 + 11) + \log_6(-15 + 6) = \log_6(-4) + \log_6(-9) \][/tex]
This is invalid since the arguments of the logarithms are negative.
Therefore, the only valid solution is:
[tex]\[ x = -2 \][/tex]
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