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

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

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

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

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Popular >

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

No Solution for y \in R

Solution steps

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

Apply log rules

y - 1 = e (3 y + 2)

Solve

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

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

Verify Solutions:

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

False

The solution is

No Solution for y \in R

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

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

lo g_{10} (5) + lo g_{10} (x) = 2

ln (x) + ln (x + 1) = 0

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

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