If `x=log_(2a)((bcd)/2), y=log_(3b)((acd)/3), z=log_(4c)((abd)/4) and w=log_(5d)((abc)/5)` and `1/(x+1)+1/(y+1)+1/(z+1)+1/(w+1) = log_(abcd)N+1,` then value of `N/40` is

Options

A. 40
B. 80
C. 120
D. 160

Answer

To solve the given problem step by step, we start with the definitions of \( x, y, z, \) and \( w \):

1. **Given Definitions**:
\[ x = \log_{2a}\left(\frac{bcd}{2}\right), \quad y = \log_{3b}\left(\frac{acd}{3}\right), \quad z = \log_{4c}\left(\frac{abd}{4}\right), \quad w = \log_{5d}\left(\frac{abc}{5}\right) \]

2. **Rewriting the Expression**:
We need to evaluate:
\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} + \frac{1}{w+1} = \log_{abcd} N + 1 \]

3. **Substituting Values**:
We can rewrite \( x, y, z, \) and \( w \) using the logarithmic identity \( \log_a b + 1 = \log_a (ab) \):
\[ \frac{1}{x+1} = \frac{1}{\log_{2a}\left(\frac{bcd}{2}\right) + 1} = \frac{1}{\log_{2a}\left(\frac{bcd}{2}\right) + \log_{2a}(2a)} = \frac{1}{\log_{2a}\left(bcd\right)} = \log_{\frac{bcd}{2}}(2a) \]
Similarly, we can write:
\[ \frac{1}{y+1} = \log_{\frac{acd}{3}}(3b), \quad \frac{1}{z+1} = \log_{\frac{abd}{4}}(4c), \quad \frac{1}{w+1} = \log_{\frac{abc}{5}}(5d) \]

4. **Combining the Terms**:
Now we combine these logarithmic expressions:
\[ \log_{\frac{bcd}{2}}(2a) + \log_{\frac{acd}{3}}(3b) + \log_{\frac{abd}{4}}(4c) + \log_{\frac{abc}{5}}(5d) \]

5. **Using Logarithmic Properties**:
Using the property \( \log_a b + \log_a c = \log_a(bc) \):
\[ = \log_{abcd}\left(2a \cdot 3b \cdot 4c \cdot 5d\right) \]

6. **Calculating the Product**:
The product simplifies to:
\[ 2 \cdot 3 \cdot 4 \cdot 5 \cdot abcd = 120 \cdot abcd \]

7. **Final Expression**:
Thus, we have:
\[ \log_{abcd}(120 \cdot abcd) = \log_{abcd}(120) + 1 \]

8. **Comparing with the Given Equation**:
Now we can compare:
\[ \log_{abcd}(N) + 1 = \log_{abcd}(120) + 1 \]
This implies:
\[ N = 120 \]

9. **Finding \( \frac{N}{40} \)**:
Finally, we calculate:
\[ \frac{N}{40} = \frac{120}{40} = 3 \]

Thus, the value of \( \frac{N}{40} \) is \( \boxed{3} \).