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

The answer to solvefor x,ln(y-1)-ln(2)=x+ln(x) is x=\W(1/2 (y-1))

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

solve for

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

x = W (\frac{1}{2} (y - 1))

Solution steps

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

Switch sides

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

Move

x

to the right side

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

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n (x)} = e^{l n (y - 1) - l n (2) - x}

Simplify

e^{l n (x)} : x

Simplify

e^{l n (y - 1) - l n (2) - x} : \frac{1}{2} e^{- x} (y - 1)

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

Solve

x = \frac{1}{2} e^{- x} (y - 1) : x = W (\frac{1}{2} (y - 1))

x = W (\frac{1}{2} (y - 1))

Verify Solutions:

x = W (\frac{1}{2} (y - 1))

True

The solution is

x = W (\frac{1}{2} (y - 1))

\frac{2 x - 1}{5} = \frac{5 x - 11}{4}

b (b - x) = a (a - x)

3^{9} = 2

\frac{x}{2} + \frac{x}{4} = 5

43 = 3 - \frac{5}{9} z

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