What is the answer to x=-2log_{2}(x) ?

The answer to x=-2log_{2}(x) is x=(2\W_{0}((ln(2))/2))/(ln(2))

Solve for x: x=-2log_{2}(x)

The solution is x=(2\W_{0}((ln(2))/2))/(ln(2))

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

x = \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )}

Decimal Notation

x = 0.76666 \dots

Solution steps

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

Switch sides

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

Divide both sides by

- 2

lo g_{2} (x) = - \frac{x}{2}

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

2^{l o g_{2} (x)} = 2^{- \frac{x}{2}}

Simplify

2^{l o g_{2} (x)} : x

x = 2^{- \frac{x}{2}}

Solve

x = 2^{- \frac{x}{2}} : x = \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )}

x = \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )}

Verify Solutions:

x = \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )}

True

The solution is

x = \frac{2 W _{0} ( \frac{l n ( 2 )}{2} )}{ln ( 2 )}

x, y = - lo g_{10} (x)

lo g_{10} (x - \frac{2}{4}) = 2

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

lo g_{5} (x) = 1 + lo g_{x} (25)

lo g_{x} (8 x - 15) = 2

What is the answer to x=-2log_{2}(x) ?
The answer to x=-2log_{2}(x) is x=(2\W_{0}((ln(2))/2))/(ln(2))
Solve for x: x=-2log_{2}(x)
The solution is x=(2\W_{0}((ln(2))/2))/(ln(2))