If x and y are positive real numbers such that \log(x + y) = \log x + \log y, which of the… — ACT

If $x$ and $y$ are positive real numbers such that $\log(x + y) = \log x + \log y$ , which of the following equations expresses $y$ in terms of $x$ ?

A
$y = \frac{x}{x+1}$
B
$y = \frac{x}{1-x}$
$y = \frac{x}{x-1}$ (where $x > 1$ )Answer
D
$y = x$
E
Any positive real number $y$ , because the equation is a general identity for all positive $x$ and $y$

y = \frac{x}{x-1}

(where

x > 1

)

By applying the product property of logarithms, the right side of the given equation

\log x + \log y

simplifies to

\log(xy)

. This transforms the equation to

\log(x+y) = \log(xy)

. Equating the arguments gives the algebraic equation

x+y=xy

. Rearranging this equation to group all

y

terms on one side yields

x = xy - y

. Factoring out

y

gives

x = y(x-1)

. Finally, dividing both sides by

x-1

results in the equation stating that

y

is equal to

x

divided by the quantity

x

minus one.

Apply the product property of logarithms to the right side of the equation:

\log x + \log y = \log(xy)

The equation becomes

\log(x + y) = \log(xy)

This allows both sides to be expressed as a single logarithm with the same base.

Equate the arguments of the logarithms, since

\log(u) = \log(v)

implies

u = v

for positive real numbers.

x + y = xy

To eliminate the logarithms and work with a purely algebraic equation.

Isolate the terms containing

y

on one side of the equation.

x = xy - y

Grouping the

y

terms on the same side is a necessary step to factor out the variable we want to solve for.

Factor out

y

from the right side and divide both sides by the remaining coefficient

x - 1

x = y(x - 1) \implies y = \frac{x}{x-1}

(for

x > 1

to keep

y

positive and avoid division by zero).

This completely isolates

y

in terms of

x

Properties of Logarithms and Algebraic Isolation

Question