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

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

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

solve for

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

y = e^{x - 2} + 1

Solution steps

x = ln (y - 1) + 2

Switch sides

ln (y - 1) + 2 = x

Move

2

to the right side

ln (y - 1) = x - 2

Apply log rules

y - 1 = e^{x - 2}

Solve

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

y = e^{x - 2} + 1

Verify Solutions:

y = e^{x - 2} + 1

True

The solution is

y = e^{x - 2} + 1

lo g_{10} (x) + lo g_{10} (x + 3) = 6

lo g_{10} (x) + lo g_{10} (x + 3) = 5

lo g_{10} (2 x - 1) = 3 x

lo g_{6} (4 x - 5) = 1

lo g_{x + 5} (2 x^{2} + 5 x + 1) = 2

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