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

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

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solvefor x,ln(x-y)+k=ln(z+1)

solve for

x, ln (x - y) + k = ln (z + 1)

x = e^{- k} z + e^{- k} + y

Solution steps

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

Move

k

to the right side

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

Apply log rules

x - y = e^{- k} z + e^{- k}

Solve

x - y = e^{- k} z + e^{- k} : x = e^{- k} z + e^{- k} + y

x = e^{- k} z + e^{- k} + y

lo g_{2} ((2 x + 4)^{7}) = 21

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

\frac{1}{2} = lo g_{4} (2 x - 1)

- 1 - ln (x) = 0

x, 0.6 = lo g_{10} (x)

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