The problem asks us to find the value of a real number '$n$' such that $n > 1$ and it satisfies the given logarithmic equation:

$ \frac{1}{\log_2n} + \frac{1}{\log_3n} + \frac{1}{\log_4n} = 1 $

Understanding the Logarithm Equation

We are given an equation involving sums of reciprocals of logarithms. To solve for '$n$', we need to simplify this equation using properties of logarithms.

Applying Logarithm Properties

A key property of logarithms is the change of base rule, which implies:

$ \frac{1}{\log_b a} = \log_a b $

Using this property, we can rewrite each term in the given equation:

• $ \frac{1}{\log_2n} = \log_n 2 $
• $ \frac{1}{\log_3n} = \log_n 3 $
• $ \frac{1}{\log_4n} = \log_n 4 $

Substituting these back into the original equation, we get:

$ \log_n 2 + \log_n 3 + \log_n 4 = 1 $

Combining Logarithmic Terms

Another important logarithm property is the sum of logarithms with the same base: $ \log_b x + \log_b y = \log_b (xy) $. Applying this property to the left side of our equation:

$ \log_n (2 \times 3 \times 4) = 1 $

Calculating the Product

First, calculate the product inside the logarithm:

$ 2 \times 3 \times 4 = 6 \times 4 = 24 $

So the equation simplifies to:

$ \log_n 24 = 1 $

Solving for n

The definition of a logarithm states that if $ \log_b a = c $, then $ b^c = a $. Applying this definition to $ \log_n 24 = 1 $:

$ n^1 = 24 $

This directly gives us the value of '$n$':

$ n = 24 $

Verification

We must check if the solution satisfies the condition given in the question, which is $ n > 1 $. Since $ 24 > 1 $, the value $ n = 24 $ is a valid solution.

Conclusion

The value of the real number '$n$' that satisfies the equation $ \frac{1}{\log_2n} + \frac{1}{\log_3n} + \frac{1}{\log_4n} = 1 $ is 24.