What is the answer to solvefor x,ln(x+1)+ln(2y)=k ?

The answer to solvefor x,ln(x+1)+ln(2y)=k is x=(e^k)/(2y)-1

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solvefor x,ln(x+1)+ln(2y)=k

solve for

x, ln (x + 1) + ln (2 y) = k

x = \frac{e ^{k}}{2 y} - 1

Solution steps

ln (x + 1) + ln (2 y) = k

Move

ln (2 y)

to the right side

ln (x + 1) = k - ln (2 y)

Apply log rules

x + 1 = \frac{e ^{k}}{2 y}

Solve

x + 1 = \frac{e ^{k}}{2 y} : x = \frac{e ^{k}}{2 y} - 1

x = \frac{e ^{k}}{2 y} - 1

lo g_{5} ((x + 2)^{2}) = 2

ln (x^{2}) = 0.456

lo g_{x} (9) = - 1

r, d = ln (1 + rs)

ln (r) = - 2

What is the answer to solvefor x,ln(x+1)+ln(2y)=k ?
The answer to solvefor x,ln(x+1)+ln(2y)=k is x=(e^k)/(2y)-1