QuestionAugust 27, 2025

Solve for x. log_(2)(x-5)=3-log_(2)(x-7) If there is more than one solution , separate them with commas. If there is no solution, click on "No solution". x=square

Solve for x. log_(2)(x-5)=3-log_(2)(x-7) If there is more than one solution , separate them with commas. If there is no solution, click on "No solution". x=square
Solve for x. log_(2)(x-5)=3-log_(2)(x-7) If there is more than one solution , separate them with commas. If there is no solution, click on "No solution". x=square

Solution
4.2(115 votes)

Answer

9 Explanation 1. Simplify the equation Combine logarithms: log_{2}(x-5) + log_{2}(x-7) = 3. 2. Use logarithm property Apply **log_{b}(a) + log_{b}(c) = log_{b}(a \cdot c)**: log_{2}((x-5)(x-7)) = 3. 3. Convert to exponential form **log_{b}(a) = c \Rightarrow a = b^c**: (x-5)(x-7) = 2^3. 4. Solve the quadratic equation Expand and solve: x^2 - 12x + 35 = 8, leading to x^2 - 12x + 27 = 0. 5. Factor the quadratic Factor: (x-9)(x-3) = 0. 6. Find solutions Solutions are x = 9 and x = 3. 7. Verify solutions Check in original equation: x = 3 is invalid as it leads to negative inside the log. Only x = 9 is valid.

Explanation

1. Simplify the equation<br /> Combine logarithms: $log_{2}(x-5) + log_{2}(x-7) = 3$.<br />2. Use logarithm property<br /> Apply **$log_{b}(a) + log_{b}(c) = log_{b}(a \cdot c)$**: $log_{2}((x-5)(x-7)) = 3$.<br />3. Convert to exponential form<br /> **$log_{b}(a) = c \Rightarrow a = b^c$**: $(x-5)(x-7) = 2^3$.<br />4. Solve the quadratic equation<br /> Expand and solve: $x^2 - 12x + 35 = 8$, leading to $x^2 - 12x + 27 = 0$.<br />5. Factor the quadratic<br /> Factor: $(x-9)(x-3) = 0$.<br />6. Find solutions<br /> Solutions are $x = 9$ and $x = 3$.<br />7. Verify solutions<br /> Check in original equation: $x = 3$ is invalid as it leads to negative inside the log. Only $x = 9$ is valid.
Click to rate:

Similar Questions