What is the answer to log_{e}(3x+5)+log_{e}(2)=2 ?

The answer to log_{e}(3x+5)+log_{e}(2)=2 is x=(e^2-10)/6

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log_{e}(3x+5)+log_{e}(2)=2

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

x = \frac{e ^{2} - 10}{6}

Decimal Notation

x = - 0.43515 \dots

Solution steps

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

Move

lo g_{e} (2)

to the right side

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

Apply log rules

3 x + 5 = \frac{e ^{2}}{2}

Solve

3 x + 5 = \frac{e ^{2}}{2} : x = \frac{e ^{2} - 10}{6}

x = \frac{e ^{2} - 10}{6}

Verify Solutions:

x = \frac{e ^{2} - 10}{6}

True

The solution is

x = \frac{e ^{2} - 10}{6}

lo g_{3} ((4 x + 2)^{3}) = 6

ln (x + e^{2}) = 0

lo g_{3} (X + 2) = - 2

a, ln (a) = 2 ln (b)

2 = lo g_{2} (y)

What is the answer to log_{e}(3x+5)+log_{e}(2)=2 ?
The answer to log_{e}(3x+5)+log_{e}(2)=2 is x=(e^2-10)/6