What is the answer to 3log_{e}(x+1)=-log_{e}(2x-1) ?

The answer to 3log_{e}(x+1)=-log_{e}(2x-1) is x=(sqrt(5)-1)/2

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3log_{e}(x+1)=-log_{e}(2x-1)

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

x = \frac{5 - 1}{2}

Solution steps

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

Add

lo g_{e} (2 x - 1)

to both sides

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

Simplify

3 lo g_{e} (x + 1) + lo g_{e} (2 x - 1) = 0

Apply log rules

(x + 1)^{3} (2 x - 1) = 1

Solve

(x + 1)^{3} (2 x - 1) = 1 : x = - \frac{5 + 1}{2}, x = \frac{5 - 1}{2}

x = - \frac{5 + 1}{2}, x = \frac{5 - 1}{2}

Verify Solutions:

x = - \frac{5 + 1}{2}

False

, x = \frac{5 - 1}{2}

True

The solution is

x = \frac{5 - 1}{2}

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

4 = \frac{ln ( 4 x )}{ln ( 2 x )}

2 lo g_{2} (x) = x

lo g_{10} (3 + x) = lo g_{10} (x) + 2

lo g_{2} (z + 6) = lo g_{2} (z - 1) + 3

What is the answer to 3log_{e}(x+1)=-log_{e}(2x-1) ?
The answer to 3log_{e}(x+1)=-log_{e}(2x-1) is x=(sqrt(5)-1)/2