Solve: $log (x + y) = log x y + a$

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Solution

$ln (x + y) = ln x y + a$
$\Rightarrow ln (x + y) - ln x y = a$
$\Rightarrow ln (\frac{x + y}{x y}) = a$
$\Rightarrow \frac{1}{x} + \frac{1}{y} = e^{a}$

Hence, this is the answer.

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Similar questions

Solve for x and y; if $x > 0$ and $y > 0$ :
$l o g x y = l o g \frac{x}{y} + 2 l o g 2 = 2$ .

Given : $\frac{l o g x}{l o g y} = \frac{3}{2}$ and $l o g (x y) = 5$ ; find the values of $x$ and $y$ .

log x y = log \frac{x}{y} + 2 log 2 = 2

x > 0

y > 0

x

y

\frac{log x}{log y} = \frac{3}{2}

log (x y) = 5

x \times y

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