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

The answer to solvefor y,ln(y-1)-2=x+ln(x) is y=e^{x+2}x+1 {x>0}

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

solve for

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

y = e^{x + 2} x + 1 {x > 0}

Solution steps

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

Move

2

to the right side

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

Apply log rules

y - 1 = e^{x + 2} x

Solve

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

y = e^{x + 2} x + 1

Verify Solutions:

y = e^{x + 2} x + 1 {x > 0}

The solution is

y = e^{x + 2} x + 1 {x > 0}

ln (x + 2) = ln (7 x)

ln (3 x + 6) = 5

ln (1 - x) + ln (3 - x) = ln (8)

4 lo g_{4} (2) = lo g_{4} (x - 2)

lo g_{10} (x) + lo g_{10} (2) = 1

What is the answer to solvefor y,ln(y-1)-2=x+ln(x) ?
The answer to solvefor y,ln(y-1)-2=x+ln(x) is y=e^{x+2}x+1 {x>0}