Question

If \(\int _0^k\frac{1}{1+4^x}dx = log_2(\frac{4}{3})\), then \(k =\) _____

• A. \(\frac{2}{\sqrt{3}}\)
• B. \(\frac{3}{2}\)
• C. \(\frac{\sqrt{3}}{2}\)
• D. \(\frac{2}{3}\)

Hint:

Write 1/(1+4^x) = 1 - 4^x/(1+4^x) and integrate.

Answer 1

The Correct Option is B

Step 1: Rewrite the integrand:
\(\dfrac{1}{1+4^x}=1-\dfrac{4^x}{1+4^x}\).

Step 2: Integrate:
\(\int_0^k\dfrac{4^x}{1+4^x}dx=\dfrac{1}{\ln4}\left[\ln(1+4^x)\right]_0^k=\dfrac{1}{\ln4}\ln\dfrac{1+4^k}{2}\). So \(\int_0^k\dfrac{dx}{1+4^x}=k-\log_4\dfrac{1+4^k}{2}\).

Step 3: Set it equal to log base 2 of 4/3:
\(\log_2\dfrac43=\log_4\dfrac{16}{9}\). So \(k-\log_4\dfrac{1+4^k}{2}=\log_4\dfrac{16}{9}\), which gives \(\log_4\dfrac{4^k\cdot2}{1+4^k}=\log_4\dfrac{16}{9}\).

Step 4: Solve:
\(\dfrac{2\cdot4^k}{1+4^k}=\dfrac{16}{9}\). Cross-multiply: \(18\cdot4^k=16+16\cdot4^k\), so \(2\cdot4^k=16\), \(4^k=8\), and \(k=\dfrac32\). Option B.

Step 5: Why the other options are wrong.
For \(k=\tfrac{2}{\sqrt3}\), \(\tfrac{\sqrt3}{2}\), \(\tfrac23\), \(4^k\) is not 8, so the equation fails.

Final Answer:
k = 3/2. \[ \boxed{\text{(B) }\dfrac32} \]