Question

For x > 1, if $(2x)^{2y} = 4e^{2x-2y}$, then $(1 + \log_e 2x)^2 \frac{dy}{dx}$ is equal to

• A. $\frac{x \log_e 2x + \log_e 2}{x}$
• B. $\frac{x \log_e 2x - \log_e 2}{x}$
• C. $x \log_e 2x + \frac{\log_e 2}{x}$
• D. $x \log_e 2x - \frac{\log_e 2}{2}$

Answer 1

Answer: (B)

$\frac{x \log_e 2x - \log_e 2}{x}$

Answer 2

Given

$$(2x)^{2y} = 4e^{2x-2y}.$$

1. Taking logarithms on both sides:

   $$\ln\left((2x)^{2y}\right) = \ln\left(4e^{2x-2y}\right).$$

   This gives:

   $$2y \ln(2x) = \ln4 + 2x - 2y.$$

   Since $\ln4 = 2\ln2$, divide the equation by 2:

   $$y\ln(2x) = \ln2 + x - y.$$

2. Rearranging:

   $$y\ln(2x) + y = x + \ln2 \quad\Longrightarrow\quad y\big(\ln(2x)+1\big) = x + \ln2.$$

   Therefore:

   $$y = \frac{x+\ln2}{\ln(2x)+1}.$$

3. Differentiate $y$ with respect to $x$ using the quotient rule:

   $$\frac{dy}{dx} = \frac{(1)(\ln(2x)+1) - (x+\ln2)\left(\frac{1}{x}\right)}{\big(\ln(2x)+1\big)^2}.$$

   Simplify the numerator:

   $$\ln(2x)+1 - \left(1 + \frac{\ln2}{x}\right) = \ln(2x) - \frac{\ln2}{x}.$$

   Thus,

   $$\frac{dy}{dx} = \frac{\ln(2x) - \frac{\ln2}{x}}{\big(\ln(2x)+1\big)^2}.$$

4. Multiply by $(1+\ln(2x))^2$:

   $$(1+\ln(2x))^2 \frac{dy}{dx} = \ln(2x) - \frac{\ln2}{x}.$$

   Rewriting,

   $$\ln(2x) - \frac{\ln2}{x} = \frac{x\ln(2x)-\ln2}{x}.$$