11 values of x. log_(2)(x+2)+log_(2)(x+6)=5

Combine logarithmic expressions: Apply the product rule of logarithms to combine the two logarithmic expressions. The product rule of logarithms states that log_b(a) + log_b(c) = log_b(a*c), where b is the base of the logarithms. log₂(x+2) + log₂(x+6) = log₂((x+2)(x+6))Set equal and rewrite: Set the combined logarithm equal to 5 and rewrite the equation in exponential form. log₂((x+2)(x+6)) = 5 can be rewritten as 2^5 = (x+2)(x+6)Calculate and expand: Calculate 2^5 and expand the right side of the equation. 2^5 = 32 (x+2)(x+6) = x^2 + 6x + 2x + 12 x^2 + 8x + 12 = 32Subtract and solve for x: Subtract 32 from both sides of the equation to set it to zero and solve for x. x^2 + 8x + 12 - 32 = 0 x^2 + 8x - 20 = 0Factor the quadratic: Factor the quadratic equation. We need to find two numbers that multiply to -20 and add up to 8. These numbers are 10 and -2. (x + 10)(x - 2) = 0Solve for x: Solve for x by setting each factor equal to zero. x + 10 = 0 or x - 2 = 0 x = -10 or x = 2Check for extraneous solutions: Check for extraneous solutions by substituting the values of x back into the original logarithmic equation. For x = -10: log₂(-10+2) + log₂(-10+6) is undefined because logarithms of negative numbers are not real. For x = 2: log₂(2+2) + log₂(2+6) = log₂(4) + log₂(8) = 2 + 3 = 5, which is true.

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Find the $11$ values of $x$ . $\newline$ $\log_{2}(x+2)+\log_{2}(x+6)=5$

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Precalculus

Quotient property of logarithms

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Q. Find the

11

values of

x

\newline

\log_{2}(x+2)+\log_{2}(x+6)=5

Combine logarithmic expressions: Apply the product rule of logarithms to combine the two logarithmic expressions. $\newline$ The product rule of logarithms states that $\log_b(a) + \log_b(c) = \log_b(a*c)$ , where $b$ is the base of the logarithms. $\newline$ $\log_2(x+2) + \log_2(x+6) = \log_2((x+2)(x+6))$
Set equal and rewrite: Set the combined logarithm equal to $5$ and rewrite the equation in exponential form. $\log_2((x+2)(x+6)) = 5$ can be rewritten as $2^5 = (x+2)(x+6)$
Calculate and expand: Calculate $2^5$ and expand the right side of the equation. $\newline$ $2^5 = 32$ $\newline$ $(x+2)(x+6) = x^2 + 6x + 2x + 12$ $\newline$ $x^2 + 8x + 12 = 32$
Subtract and solve for x: Subtract $32$ from both sides of the equation to set it to zero and solve for $x$ . $\newline$ $x^2 + 8x + 12 - 32 = 0$ $\newline$ $x^2 + 8x - 20 = 0$
Factor the quadratic: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $-20$ and add up to $8$ . These numbers are $10$ and $-2$ . $\newline$ $(x + 10)(x - 2) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x + 10 = 0$ or $x - 2 = 0$ $\newline$ $x = -10$ or $x = 2$
Check for extraneous solutions: Check for extraneous solutions by substituting the values of $x$ back into the original logarithmic equation. $\newline$ For $x = -10$ : $\newline$ $\log_2(-10+2) + \log_2(-10+6)$ is undefined because logarithms of negative numbers are not real. $\newline$ For $x = 2$ : $\newline$ $\log_2(2+2) + \log_2(2+6) = \log_2(4) + \log_2(8) = 2 + 3 = 5$ , which is true.