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To determine the value of [tex]\( x \)[/tex] that satisfies the equation [tex]\(\log _4(x+14) + \log _4(x+8) = 2\)[/tex], follow these steps:
1. Combine the logarithmic terms:
Recall the logarithm property that states [tex]\(\log_b A + \log_b B = \log_b (A \cdot B)\)[/tex]. We can combine the logarithms on the left side of the equation:
[tex]\[ \log_4(x+14) + \log_4(x+8) = \log_4\left((x+14)(x+8)\right) \][/tex]
So the equation becomes:
[tex]\[ \log_4\left((x+14)(x+8)\right) = 2 \][/tex]
2. Exponentiate both sides to remove the logarithm:
Exponentiating both sides with base 4 gives:
[tex]\[ (x+14)(x+8) = 4^2 \][/tex]
Simplifying the right side, we get:
[tex]\[ (x+14)(x+8) = 16 \][/tex]
3. Expand and simplify the quadratic equation:
Distribute the terms on the left side:
[tex]\[ x^2 + 8x + 14x + 112 = 16 \][/tex]
Combine like terms:
[tex]\[ x^2 + 22x + 112 = 16 \][/tex]
4. Set the equation to zero:
Subtract 16 from both sides to form a standard quadratic equation:
[tex]\[ x^2 + 22x + 96 = 0 \][/tex]
5. Solve the quadratic equation:
To solve for [tex]\( x \)[/tex], we can either factor the quadratic equation or use the quadratic formula. Let's try factoring first:
[tex]\[ x^2 + 22x + 96 = (x + 6)(x + 16) = 0 \][/tex]
Set each factor equal to zero:
[tex]\[ x + 6 = 0 \quad \text{or} \quad x + 16 = 0 \][/tex]
Solving these, we get:
[tex]\[ x = -6 \quad \text{or} \quad x = -16 \][/tex]
6. Check the solutions in the original logarithmic equation:
We must check that our solutions are valid by substituting back into the original equation. Logarithms are only defined for positive arguments.
- For [tex]\( x = -6 \)[/tex]:
[tex]\[ \log_4(-6 + 14) + \log_4(-6 + 8) = \log_4(8) + \log_4(2) \][/tex]
Note that both [tex]\(\log_4(8)\)[/tex] and [tex]\(\log_4(2)\)[/tex] are defined and positive.
- For [tex]\( x = -16 \)[/tex]:
[tex]\[ \log_4(-16 + 14) + \log_4(-16 + 8) = \log_4(-2) + \log_4(-8) \][/tex]
Both [tex]\(\log_4(-2)\)[/tex] and [tex]\(\log_4(-8)\)[/tex] are undefined because the arguments are negative.
Therefore, [tex]\( x = -16 \)[/tex] is not a valid solution.
Thus, the value of [tex]\( x \)[/tex] that satisfies the equation is:
[tex]\[ \boxed{-6} \][/tex]
1. Combine the logarithmic terms:
Recall the logarithm property that states [tex]\(\log_b A + \log_b B = \log_b (A \cdot B)\)[/tex]. We can combine the logarithms on the left side of the equation:
[tex]\[ \log_4(x+14) + \log_4(x+8) = \log_4\left((x+14)(x+8)\right) \][/tex]
So the equation becomes:
[tex]\[ \log_4\left((x+14)(x+8)\right) = 2 \][/tex]
2. Exponentiate both sides to remove the logarithm:
Exponentiating both sides with base 4 gives:
[tex]\[ (x+14)(x+8) = 4^2 \][/tex]
Simplifying the right side, we get:
[tex]\[ (x+14)(x+8) = 16 \][/tex]
3. Expand and simplify the quadratic equation:
Distribute the terms on the left side:
[tex]\[ x^2 + 8x + 14x + 112 = 16 \][/tex]
Combine like terms:
[tex]\[ x^2 + 22x + 112 = 16 \][/tex]
4. Set the equation to zero:
Subtract 16 from both sides to form a standard quadratic equation:
[tex]\[ x^2 + 22x + 96 = 0 \][/tex]
5. Solve the quadratic equation:
To solve for [tex]\( x \)[/tex], we can either factor the quadratic equation or use the quadratic formula. Let's try factoring first:
[tex]\[ x^2 + 22x + 96 = (x + 6)(x + 16) = 0 \][/tex]
Set each factor equal to zero:
[tex]\[ x + 6 = 0 \quad \text{or} \quad x + 16 = 0 \][/tex]
Solving these, we get:
[tex]\[ x = -6 \quad \text{or} \quad x = -16 \][/tex]
6. Check the solutions in the original logarithmic equation:
We must check that our solutions are valid by substituting back into the original equation. Logarithms are only defined for positive arguments.
- For [tex]\( x = -6 \)[/tex]:
[tex]\[ \log_4(-6 + 14) + \log_4(-6 + 8) = \log_4(8) + \log_4(2) \][/tex]
Note that both [tex]\(\log_4(8)\)[/tex] and [tex]\(\log_4(2)\)[/tex] are defined and positive.
- For [tex]\( x = -16 \)[/tex]:
[tex]\[ \log_4(-16 + 14) + \log_4(-16 + 8) = \log_4(-2) + \log_4(-8) \][/tex]
Both [tex]\(\log_4(-2)\)[/tex] and [tex]\(\log_4(-8)\)[/tex] are undefined because the arguments are negative.
Therefore, [tex]\( x = -16 \)[/tex] is not a valid solution.
Thus, the value of [tex]\( x \)[/tex] that satisfies the equation is:
[tex]\[ \boxed{-6} \][/tex]
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