Question

Find $log_x⁵ y ⋅ log_y x²,$ where $x, y > 0$.

a) $10logₓ y$
b) $5logₓ y$
c) $2logₓ y$
d) $0.5logₓ y$

Answer 1

The expression log_x⁵ y ⋅ log_y x² simplifies to 10 ⋅ log_x y using properties of logarithms and the change of base formula. The correct option is (a) 10logₓ y.

Step-by-step explanation:

To find log_x⁵ y ⋅ log_y x², we use the property that the logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number. Therefore:

log_x y⁵ = 5 ⋅ log_x y

log_y x² = 2 ⋅ log_y x

Now, multiplying these two expressions, we get:

(5 ⋅ log_x y) ⋅ (2 ⋅ log_y x) = 10 ⋅ log_x y ⋅ log_y x

Using the change of base formula, which states that log_a b ⋅ log_b a = 1, the expression simplifies:

10 ⋅ log_x y ⋅ log_y x = 10 ⋅ (1)

Therefore, the answer is:

10 ⋅ log_x y

Which corresponds to option (a).