Identify the solution set of 6 ln e = eln 2x {2} {3} {6}

The Correct Answer and Explanation is:

To solve the equation: 6ln⁡e=eln⁡2×6 \ln e = e^{\ln 2x}

Step 1: Simplify the left side

We use the identity: ln⁡e=1\ln e = 1

So, 6ln⁡e=6⋅1=66 \ln e = 6 \cdot 1 = 6

Step 2: Simplify the right side

Use the identity: eln⁡a=a, for a>0e^{\ln a} = a, \text{ for } a > 0

So: eln⁡2x=2xe^{\ln 2x} = 2x

Step 3: Solve the equation

Now the equation is: 6=2×6 = 2x

Divide both sides by 2: x=3x = 3

Step 4: Check domain restrictions

Since we used ln⁡2x\ln 2x, and logarithms are only defined for positive arguments, we must have: 2x>0⇒x>02x > 0 \Rightarrow x > 0

Our solution x=3x = 3 satisfies this condition.

✅ Final Answer:

The solution set is {3}\{3\}.

📘 Explanation

The given equation is 6ln⁡e=eln⁡2×6 \ln e = e^{\ln 2x}. To find the solution, we simplify both sides using logarithmic and exponential identities. The natural logarithm of ee, or ln⁡e\ln e, equals 1, so the left-hand side simplifies to 66.

On the right-hand side, we see eln⁡2xe^{\ln 2x}, which seems complicated at first glance. However, we apply a useful identity from logarithmic and exponential functions: for any positive real number aa, eln⁡a=ae^{\ln a} = a. This is because exponentiation and logarithms are inverse operations.

Applying this identity, eln⁡2x=2xe^{\ln 2x} = 2x, as long as 2x>02x > 0, meaning x>0x > 0. This is important because the logarithm is only defined for positive arguments, so 2x2x must be positive.

So now, our simplified equation is: 6=2×6 = 2x

To isolate xx, divide both sides by 2: x=3x = 3

Finally, we confirm that this value satisfies all domain requirements — specifically, x=3>0x = 3 > 0, so 2x=6>02x = 6 > 0, and ln⁡2x\ln 2x is defined.

Thus, the only valid solution is x=3x = 3, and the solution set is {3}\{3\}. The other options ({2}, {6}) do not satisfy the equation.