Questions: Solve for (x) : [ log (x)+log (x+4)=1 x=square ]

Transcript text: Solve for $x$ : \[ \begin{array}{l} \log (x)+\log (x+4)=1 \\ x=\square \end{array} \]

Solution

Solution Steps

To solve the equation $\log(x) + \log(x+4) = 1$, we can use the properties of logarithms. Specifically, we can use the product rule for logarithms, which states that $\log(a) + \log(b) = \log(ab)$. This allows us to combine the logarithms into a single logarithm. Then, we can exponentiate both sides to solve for $x$.

Solution Approach

Use the product rule for logarithms to combine the two logarithms into one: $\log(x) + \log(x+4) = \log(x(x+4))$.
Set the combined logarithm equal to 1: $\log(x(x+4)) = 1$.
Exponentiate both sides to remove the logarithm: $x(x+4) = 10^1$.
Solve the resulting quadratic equation for $x$.

Step 1: Combine the Logarithms

Using the product rule for logarithms, we combine the two logarithms: \[ \log(x) + \log(x + 4) = \log(x(x + 4)) \] Thus, the equation becomes: \[ \log(x(x + 4)) = 1 \]

Step 2: Exponentiate Both Sides

Exponentiate both sides to remove the logarithm: \[ x(x + 4) = 10^1 \] This simplifies to: \[ x^2 + 4x = 10 \]

Step 3: Solve the Quadratic Equation

Rearrange the equation to standard quadratic form: \[ x^2 + 4x - 10 = 0 \] Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 4$, and $c = -10$: \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \] \[ x = \frac{-4 \pm \sqrt{16 + 40}}{2} \] \[ x = \frac{-4 \pm \sqrt{56}}{2} \] \[ x = \frac{-4 \pm 2\sqrt{14}}{2} \] \[ x = -2 \pm \sqrt{14} \]

Step 4: Filter Real Solutions

Since $x$ must be positive (as the logarithm of a non-positive number is undefined), we discard the negative solution: \[ x = -2 + \sqrt{14} \]

Step 5: Numerical Approximation

Approximating the value of $x$ to four significant digits: \[ x \approx 0.591964858646630 \approx 0.5920 \]