Question

If $P = \log_5(\log_5 3)$ and $3C + 5 - P = 405$ then C is equal to

• A. 3
• B. 4
• C. 81

Answer 1

C = (400 + log₅(log₅3)) / 3 ≈ 133.25

Answer 2

We are given $P = \log_5(\log_5 3)$ and $3C + 5 - P = 405$. We need to find the value of C.

1. From the second equation, we can isolate $3C$:

   $$3C = 405 - 5 + P = 400 + P$$

2. Now, divide by 3 to solve for $C$:

   $$C = \frac{400 + P}{3}$$

3. Substitute $P = \log_5(\log_5 3)$:

   $$C = \frac{400 + \log_5(\log_5 3)}{3}$$

4. Approximate the value of $\log_5 3$:

   $$\log_5 3 = \frac{\ln 3}{\ln 5} \approx \frac{1.0986}{1.6094} \approx 0.6826$$

5. Approximate the value of $P = \log_5(0.6826)$:

   $$P = \log_5(0.6826) = \frac{\ln 0.6826}{\ln 5} \approx \frac{-0.382}{1.6094} \approx -0.237$$

6. Substitute the approximate value of $P$ into the equation for $C$:

   $$C \approx \frac{400 - 0.237}{3} \approx \frac{399.763}{3} \approx 133.254$$

Therefore, $C \approx 133.25$. None of the given options (3, 4, 81) match this computed value.