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

The answer to ln(x)=1+ln(x-2) is x=(2e)/(e-1)

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

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

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

Solution steps

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

Apply log rules

x = e (x - 2)

Solve

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

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

Verify Solutions:

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

True

The solution is

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

lo g_{2} (x + 3) + lo g_{2} (x - 4) = 3

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

y, lo g_{b} (y) = x

lo g_{10} (3 a) = 0

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

What is the answer to ln(x)=1+ln(x-2) ?
The answer to ln(x)=1+ln(x-2) is x=(2e)/(e-1)
Solve for x: ln(x)=1+ln(x-2)
The solution is x=(2e)/(e-1)