Question

Let $n = 75600$, then find the value of $\frac{4}{log_2 n} + \frac{3}{log_3 n} + \frac{2}{log_5 n} + \frac{1}{log_7 n}$

Answer 1

1

Answer 2

Let the given expression be $E$. We are given $n = 75600$. We first find the prime factorization of $n$. $n = 75600 = 756 \times 100$ $756 = 2 \times 378 = 2^2 \times 189 = 2^2 \times 3 \times 63 = 2^2 \times 3 \times 3^2 \times 7 = 2^2 \times 3^3 \times 7^1$ $100 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2$ So, $n = (2^2 \times 3^3 \times 7^1) \times (2^2 \times 5^2) = 2^{2+2} \times 3^3 \times 5^2 \times 7^1 = 2^4 \times 3^3 \times 5^2 \times 7^1$.

The given expression is $E = \frac{4}{log_2 n} + \frac{3}{log_3 n} + \frac{2}{log_5 n} + \frac{1}{log_7 n}$. Using the change of base formula for logarithms, $\frac{1}{log_b a} = log_a b$, we can rewrite each term: $\frac{1}{log_2 n} = log_n 2$ $\frac{1}{log_3 n} = log_n 3$ $\frac{1}{log_5 n} = log_n 5$ $\frac{1}{log_7 n} = log_n 7$

Substitute these into the expression $E$: $E = 4 \times log_n 2 + 3 \times log_n 3 + 2 \times log_n 5 + 1 \times log_n 7$

Using the logarithm property $k \times log_b a = log_b a^k$, we get: $E = log_n 2^4 + log_n 3^3 + log_n 5^2 + log_n 7^1$

Using the logarithm property $log_b x + log_b y + log_b z + ... = log_b (x \times y \times z \times ...)$, we can combine the terms: $E = log_n (2^4 \times 3^3 \times 5^2 \times 7^1)$

From the prime factorization of $n$, we know that $n = 2^4 \times 3^3 \times 5^2 \times 7^1$. So, the expression becomes: $E = log_n n$

Using the logarithm property $log_b b = 1$, we have: $E = 1$