QuestionWhat is the solution to $2 \ln x=4 \ln 2 ?$ $x=$

Studdy Solution

STEP 1

What is this asking? This problem is asking us to find the value of $x$ that makes the equation $2 \ln x = 4 \ln 2$ true.
It's all about those logarithms! Watch out! Remember those logarithm rules!
It's easy to mess them up if you're not careful.

STEP 2

1. Simplify the equation using logarithm properties.
2. Isolate $x$ .

STEP 3

We can rewrite the equation using the power rule of logarithms, which says $a \cdot \ln(b) = \ln(b^a)$ .
So, let's move those coefficients inside the logarithms: $2 \ln x = \ln(x^2)$ $4 \ln 2 = \ln(2^4)$ So our equation becomes: $\ln(x^2) = \ln(2^4)$

STEP 4

We know $2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 16$ , so the equation is now: $\ln(x^2) = \ln(16)$

STEP 5

To get $x$ by itself, we can exponentiate both sides of the equation using e as the base.
This is the inverse operation of taking the natural logarithm, and it's allowed since the logarithm of a negative number is undefined, meaning $x$ can't be negative. $e^{\ln(x^2)} = e^{\ln(16)}$ Since $e^{\ln(a)} = a$ , this simplifies to: $x^2 = 16$

STEP 6

Now, we take the square root of both sides to solve for $x$ : $\sqrt{x^2} = \sqrt{16}$ This gives us two possible solutions: $x = 4 \quad \text{or} \quad x = -4$

STEP 7

Remember, we can't take the logarithm of a negative number!
Since our original equation contained $\ln(x)$ , we must have $x > 0$ .
Therefore, $x = -4$ is an extraneous solution, and we discard it.
So, our final answer is $x = 4$ .

STEP 8

$x = 4$

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