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${:[x=17],[log_(2)(2x)+log_(2)(x-7)=log_(2)(4x)]:}$

$\begin{array}{c}x=17 \\ \log _{2}(2 x)+\log _{2}(x-7)=\log _{2}(4 x)\end{array}$

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Quotient property of logarithms

Full solution

Q.

\begin{array}{c}x=17 \\ \log _{2}(2 x)+\log _{2}(x-7)=\log _{2}(4 x)\end{array}

Combine Logs: Combine the left side using the product rule for logarithms: $\log_2(a) + \log_2(b) = \log_2(ab)$ . $\log_2(2x) + \log_2(x - 7) = \log_2(2x \cdot (x - 7))$ .
Simplify Left Side: Now we have $\log_2(2x * (x - 7)) = \log_2(4x)$ . Simplify the left side: $2x * (x - 7) = 2x^2 - 14x$ .
Set Equations Equal: Since the bases of the logarithms are the same, we can set the arguments equal to each other: $2x^2 - 14x = 4x$ .
Subtract and Simplify: Subtract $4x$ from both sides to get a quadratic equation: $2x^2 - 14x - 4x = 0$ . This simplifies to $2x^2 - 18x = 0$ .
Factor Common Factor: Factor out the common factor of $2x$ : $2x(x - 9) = 0$ .
Set Factors Equal: Set each factor equal to zero: $2x = 0$ or $x - 9 = 0$ .
Solve Equations: Solve each equation: $x = 0$ or $x = 9$ .
Check Solutions: Check for extraneous solutions by plugging $x$ back into the original equation. $\newline$ For $x = 0$ : $\log_2(2\cdot0) + \log_2(0 - 7)$ does not exist because log of a negative number is undefined.
Check Solutions: Check for extraneous solutions by plugging $x$ back into the original equation. $\newline$ For $x = 0$ : $\log_2(2\cdot0) + \log_2(0 - 7)$ does not exist because log of a negative number is undefined.For $x = 9$ : $\log_2(2\cdot9) + \log_2(9 - 7) = \log_2(4\cdot9)$ checks out because $\log_2(18) + \log_2(2) = \log_2(36)$ and $18 \cdot 2 = 36$ .

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