Question

Let $P = \frac{5}{\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} + \frac{1}{\log_5 x}}$ and $(120)^P = 32$, then the value of x be:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

The problem requires simplifying a logarithmic expression for P and then using an exponential equation to find the value of x.

1. Simplify the expression for P: The given expression for P is: $P = \frac{5}{\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} + \frac{1}{\log_5 x}}$ Using the change of base formula for logarithms, $\frac{1}{\log_b a} = \log_a b$, we can rewrite each term in the denominator: $\frac{1}{\log_2 x} = \log_x 2$ $\frac{1}{\log_3 x} = \log_x 3$ $\frac{1}{\log_4 x} = \log_x 4$ $\frac{1}{\log_5 x} = \log_x 5$ Substitute these back into the expression for P: $P = \frac{5}{\log_x 2 + \log_x 3 + \log_x 4 + \log_x 5}$ Now, use the product rule for logarithms, $\log_a M + \log_a N = \log_a (MN)$: $P = \frac{5}{\log_x (2 \times 3 \times 4 \times 5)}$ Calculate the product in the argument: $2 \times 3 \times 4 \times 5 = 6 \times 20 = 120$. So, the expression for P becomes: $P = \frac{5}{\log_x 120}$ Apply the change of base formula again, $\frac{1}{\log_b a} = \log_a b$: $P = 5 \log_{120} x$

2. Solve the given equation for P: The given equation is $(120)^P = 32$. To solve for P, take the logarithm with base 120 on both sides of the equation: $\log_{120}((120)^P) = \log_{120} 32$ Using the power rule for logarithms, $\log_b (M^n) = n \log_b M$, and the property $\log_b b = 1$: $P \log_{120} 120 = \log_{120} 32$ $P \times 1 = \log_{120} 32$ $P = \log_{120} 32$

3. Equate the two expressions for P and solve for x: We have two expressions for P: From step 1: $P = 5 \log_{120} x$ From step 2: $P = \log_{120} 32$ Equating them: $5 \log_{120} x = \log_{120} 32$ Apply the power rule for logarithms to the left side: $\log_{120} (x^5) = \log_{120} 32$ Since the bases of the logarithms are the same, their arguments must be equal: $x^5 = 32$ We know that $2^5 = 32$. Therefore, $x = 2$.

4. Check domain restrictions: For the original logarithmic terms $\log_b x$ to be defined, $x$ must be greater than 0 ($x > 0$). Also, for the terms $\frac{1}{\log_b x}$ to be defined, $\log_b x$ cannot be zero, which means $x \neq 1$. Our solution $x=2$ satisfies both conditions ($2 > 0$ and $2 \neq 1$).