What is the answer to solvefor u,-ln(1+u)=ln(x) ?

The answer to solvefor u,-ln(1+u)=ln(x) is u= 1/x-1 {x>0}

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solvefor u,-ln(1+u)=ln(x)

solve for

u, - ln (1 + u) = ln (x)

u = \frac{1}{x} - 1 {x > 0}

Solution steps

- ln (1 + u) = ln (x)

Divide both sides by

- 1

ln (1 + u) = - ln (x)

Apply log rules

1 + u = \frac{1}{x}

Solve

1 + u = \frac{1}{x} : u = \frac{1}{x} - 1

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

Verify Solutions:

u = \frac{1}{x} - 1 {x > 0}

The solution is

u = \frac{1}{x} - 1 {x > 0}

3 ln (8 x + 8) + 11 = 20

2 lo g_{3} (x) = 4

lo g_{10} (\frac{x}{4}) = - 2

lo g_{10} (3 x - 2) = x

8 + 5 ln (x) = 5

What is the answer to solvefor u,-ln(1+u)=ln(x) ?
The answer to solvefor u,-ln(1+u)=ln(x) is u= 1/x-1 {x>0}