What is the answer to ln(x-1)+(1/((2x-2)))=c ?

The answer to ln(x-1)+(1/((2x-2)))=c is x=(-1+2\W(-1/(2e^c)))/(2\W(-1/(2e^c)))

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ln(x-1)+(1/((2x-2)))=c

ln (x - 1) + (\frac{1}{( 2 x - 2 )}) = c

x = \frac{- 1 + 2 W ( - \frac{1}{2 e ^{c}} )}{2 W ( - \frac{1}{2 e ^{c}} )}

Solution steps

ln (x - 1) + (\frac{1}{( 2 x - 2 )}) = c

Move

\frac{1}{2 x - 2}

to the right side

ln (x - 1) = c - \frac{1}{2 x - 2}

f (x) = g (x),

then

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

e^{l n (x - 1)} = e^{c - \frac{1}{2 x - 2}}

Simplify

e^{l n (x - 1)} : x - 1

x - 1 = e^{c - \frac{1}{2 x - 2}}

Solve

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

x = \frac{- 1 + 2 W ( - \frac{1}{2 e ^{c}} )}{2 W ( - \frac{1}{2 e ^{c}} )}

Verify Solutions:

x = \frac{- 1 + 2 W ( - \frac{1}{2 e ^{c}} )}{2 W ( - \frac{1}{2 e ^{c}} )}

True

The solution is

x = \frac{- 1 + 2 W ( - \frac{1}{2 e ^{c}} )}{2 W ( - \frac{1}{2 e ^{c}} )}

x, ln (x) = 5 k

lo g_{10} (x^{2} - 21 x) = 1

k, p k = - lo g_{10} (k)

lo g_{3} (12 x - 12) = 2 + lo g_{3} (x)

20 = 3 lo g_{2} (1 + x)

What is the answer to ln(x-1)+(1/((2x-2)))=c ?
The answer to ln(x-1)+(1/((2x-2)))=c is x=(-1+2\W(-1/(2e^c)))/(2\W(-1/(2e^c)))