What is the answer to 2=ln(3y+2)+ln(4) ?

The answer to 2=ln(3y+2)+ln(4) is y=(e^2-8)/(12)

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2=ln(3y+2)+ln(4)

2 = ln (3 y + 2) + ln (4)

y = \frac{e ^{2} - 8}{12}

Decimal Notation

y = - 0.05091 \dots

Solution steps

2 = ln (3 y + 2) + ln (4)

Switch sides

ln (3 y + 2) + ln (4) = 2

Move

ln (4)

to the right side

ln (3 y + 2) = 2 - ln (4)

Apply log rules

3 y + 2 = \frac{e ^{2}}{4}

Solve

3 y + 2 = \frac{e ^{2}}{4} : y = \frac{e ^{2} - 8}{12}

y = \frac{e ^{2} - 8}{12}

Verify Solutions:

y = \frac{e ^{2} - 8}{12}

True

The solution is

y = \frac{e ^{2} - 8}{12}

ln (4 x - 3) = ln (x + 1) + ln (3)

lo g_{10} (x) + lo g_{10} (x - 9) = 2

x, ln (3 x) = R

lo g_{10} (x) = 1000000000

x, ln (x - y^{2}) = 0

What is the answer to 2=ln(3y+2)+ln(4) ?
The answer to 2=ln(3y+2)+ln(4) is y=(e^2-8)/(12)
Solve for y: 2=ln(3y+2)+ln(4)
The solution is y=(e^2-8)/(12)