For all real numbers x and y such that x > y > 0, which of the following expressions is eq… — ACT

For all real numbers $x$ and $y$ such that $x > y > 0$ , which of the following expressions is equivalent to $\log_5(x^2 - y^2) - \log_5(x - y)$ ?

The expression is equivalent to \log_5(x + y)

The correct answer is found by using the quotient property of logarithms to combine the terms:

\log_5(x^2 - y^2) - \log_5(x - y) = \log_5\left(\frac{x^2 - y^2}{x - y}\right)

. Factoring the difference of squares in the numerator yields

\log_5\left(\frac{(x-y)(x+y)}{x-y}\right)

. Canceling the common factor

(x-y)

simplifies the expression to

\log_5(x+y)

Combine the logarithms using the quotient property of logarithms:

\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)

\log_5\left(\frac{x^2 - y^2}{x - y}\right)

This simplifies the difference between two logarithms with the same base into a single logarithm.

Factor the difference of squares in the numerator:

x^2 - y^2 = (x - y)(x + y)

\log_5\left(\frac{(x - y)(x + y)}{x - y}\right)

This allows for the cancellation of common factors in the fraction.

Cancel the common factor

(x - y)

from both the numerator and the denominator.

\log_5(x + y)

Since

x > y > 0

, the term

x - y

is non-zero, making the division valid.

Quotient property of logarithms and difference of squares factoring

Instead of applying the quotient property first, one can factor the argument of the first term using the difference of squares:

\log_5(x^2 - y^2) = \log_5((x - y)(x + y))

. Next, apply the product property of logarithms to split this term:

\log_5(x - y) + \log_5(x + y)

. Substituting this back into the original expression gives

\log_5(x - y) + \log_5(x + y) - \log_5(x - y)

, which simplifies directly to

\log_5(x + y)

Estimated Time:1m 0s

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