What is the answer to ln(x-1)=1+ln(3x+2) ?

The answer to ln(x-1)=1+ln(3x+2) is No Solution for x\in\mathbb{R}

Solve for x: ln(x-1)=1+ln(3x+2)

The solution is No Solution for x\in\mathbb{R}

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ln (x - 1) = 1 + ln (3 x + 2)

No Solution for x \in R

Solution steps

ln (x - 1) = 1 + ln (3 x + 2)

Apply log rules

x - 1 = e (3 x + 2)

Solve

x - 1 = e (3 x + 2) : x = \frac{2 e + 1}{1 - 3 e}

x = \frac{2 e + 1}{1 - 3 e}

Verify Solutions:

x = \frac{2 e + 1}{1 - 3 e}

False

The solution is

No Solution for x \in R

lo g_{2} (x) + lo g_{2} (5) = 3

lo g_{10} (24 - 7 x) = 1

ln (x y) = ln (x) + ln (y)

ln (x) + x = 1

lo g_{2 x} (16) = ln (2 x)

What is the answer to ln(x-1)=1+ln(3x+2) ?
The answer to ln(x-1)=1+ln(3x+2) is No Solution for x\in\mathbb{R}
Solve for x: ln(x-1)=1+ln(3x+2)
The solution is No Solution for x\in\mathbb{R}