Solve the equation.
-$\log _ { 2 } \sqrt { x + 3 } + \log _ { 2 } \sqrt { x } = 1$
A) $\{ - 4,1 \}$
B) { }
C) $\{ - 4 \}$
D) $\{ 1 \}$

Question

Solve the equation.
-$\log _ { 2 } \sqrt { x + 3 } + \log _ { 2 } \sqrt { x } = 1$
A) $\{ - 4,1 \}$
B) {  }
C) $\{ - 4 \}$
D) $\{ 1 \}$

Quizplus · Accepted Answer

The equation can be rewritten using the property $\log_b(m) + \log_b(n) = \log_b(mn)$ as $\log_2(\sqrt{x+3} \cdot \sqrt{x}) = 1$. This simplifies to $\log_2(\sqrt{(x+3)x}) = 1$. Converting from logarithmic to exponential form gives $\sqrt{(x+3)x} = 2$. Squaring both sides results in $(x+3)x = 4$, leading to the quadratic equation $x^2 + 3x - 4 = 0$. Solving this gives the roots $x = 1$ and $x = -4$. However, $\log_2(\sqrt{x})$ is undefined for negative $x$, so $x = -4$ is not valid. Thus, the solution is $x = 1$.

Solve the Equation log⁡2x+3+log⁡2x=1\log _ { 2 } \sqrt { x + 3 } + \log _ { 2 } \sqrt { x } = 1log2​x+3​+log2​x​=1

Solve the Equation $\log _ { 2 } \sqrt { x + 3 } + \log _ { 2 } \sqrt { x } = 1$