What is the answer to 2xlog_{e}(x)+x=0 ?

The answer to 2xlog_{e}(x)+x=0 is x= 1/(sqrt(e))

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2xlog_{e}(x)+x=0

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

x = \frac{1}{e}

Decimal

x = 0.60653 \dots

Solution steps

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

Move

x

to the right side

2 x lo g_{e} (x) = - x

Move

x

to the left side

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

Factor

2 x lo g_{e} (x) + x : x (2 lo g_{e} (x) + 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 lo g_{e} (x) + 1 = 0 : x = \frac{1}{e}

Verify Solutions:

x = 0

False

, x = \frac{1}{e}

True

The solution is

x = \frac{1}{e}

lo g_{3} (- 4 x + 10) = 3

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

lo g_{9} (x) = (lo g_{3} (x)) 2, x > 1

lo g_{x - 2} (81) = 4

lo g_{5} (x) + 6 lo g_{x} (5) = 5

What is the answer to 2xlog_{e}(x)+x=0 ?
The answer to 2xlog_{e}(x)+x=0 is x= 1/(sqrt(e))