Questions: Solve for (x) : [ log (x+4)-log (x+2)=2 x= ]

Transcript text: Solve for $x$ : \[ \begin{array}{l} \log (x+4)-\log (x+2)=2 \\ x= \end{array} \]

Solution

Solution Steps

Step 1: Set Up the Equation

We start with the logarithmic equation: \[ \log(x + 4) - \log(x + 2) = 2 \] Using the property of logarithms that states $\log(a) - \log(b) = \log\left(\frac{a}{b}\right)$, we can rewrite the equation as: \[ \log\left(\frac{x + 4}{x + 2}\right) = 2 \]

Step 2: Exponentiate to Remove the Logarithm

To eliminate the logarithm, we exponentiate both sides: \[ \frac{x + 4}{x + 2} = e^2 \]

Step 3: Solve for $x$

Cross-multiplying gives us: \[ x + 4 = e^2(x + 2) \] Expanding the right side: \[ x + 4 = e^2 x + 2e^2 \] Rearranging the equation to isolate $x$: \[ x - e^2 x = 2e^2 - 4 \] Factoring out $x$: \[ x(1 - e^2) = 2e^2 - 4 \] Thus, we can solve for $x$: \[ x = \frac{2e^2 - 4}{1 - e^2} \]

Step 4: Check Validity of the Solution

We need to ensure that the arguments of the logarithms are positive:

$x + 4 > 0$ implies $x > -4$
$x + 2 > 0$ implies $x > -2$

Since $e^2 \approx 7.389$, we find: \[ 1 - e^2 < 0 \quad \text{(since $e^2 > 1$)} \] Thus, the fraction $\frac{2e^2 - 4}{1 - e^2}$ will be valid as long as $2e^2 - 4 > 0$, which is true since $e^2 > 2$.

Final Answer

The solution for $x$ is: \[ \boxed{x = \frac{2e^2 - 4}{1 - e^2}} \]

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