What is the answer to solvefor y,e^x=ln((x-y)/(xy)) ?

The answer to solvefor y,e^x=ln((x-y)/(xy)) is y= x/(e^{e^x)x+1}

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solvefor y,e^x=ln((x-y)/(xy))

solve for

y, e^{x} = ln (\frac{x - y}{x y})

y = \frac{x}{e ^{e^{x}} x + 1}

Solution steps

e^{x} = ln (\frac{x - y}{x y})

Apply log rules

e^{e^{x}} = \frac{x - y}{x y}

Multiply both sides by

x y

e^{e^{x}} x y = \frac{x - y}{x y} x y

Simplify

e^{e^{x}} x y = x - y

Solve

e^{e^{x}} x y = x - y : y = \frac{x}{e ^{e^{x}} x + 1}

y = \frac{x}{e ^{e^{x}} x + 1}

Verify Solutions:

y = \frac{x}{e ^{e^{x}} x + 1}

True

The solution is

y = \frac{x}{e ^{e^{x}} x + 1}

lo g_{10} (x^{2}) = 0

lo g_{\frac{1}{27}} (4 x + 1) = - 1

lo g_{\frac{1}{2}} (x - 3) = 0

lo g_{x} (3) = 41

lo g_{b} (16) = - 4

What is the answer to solvefor y,e^x=ln((x-y)/(xy)) ?
The answer to solvefor y,e^x=ln((x-y)/(xy)) is y= x/(e^{e^x)x+1}