What is the answer to 2x-e^2log_{e}(x)=0 ?

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

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2x-e^2log_{e}(x)=0

2 x - e^{2} lo g_{e} (x) = 0

x = e^{2}, x = - \frac{e ^{2} W _{0} ( - \frac{2}{e ^{2}} )}{2}

Decimal Notation

x = 7.38905 \dots, x = 1.50136 \dots

Solution steps

2 x - e^{2} lo g_{e} (x) = 0

Move

2 x

to the right side

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

Divide both sides by

- e^{2}

lo g_{e} (x) = \frac{2 x}{e ^{2}}

f (x) = g (x),

then

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

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

Simplify

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

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

Solve

x = e^{\frac{2 x}{e ^{2}}} : x = e^{2}, x = - \frac{e ^{2} W _{0} ( - \frac{2}{e ^{2}} )}{2}

x = e^{2}, x = - \frac{e ^{2} W _{0} ( - \frac{2}{e ^{2}} )}{2}

Verify Solutions:

x = e^{2}

True

, x = - \frac{e ^{2} W _{0} ( - \frac{2}{e ^{2}} )}{2}

True

The solutions are

x = e^{2}, x = - \frac{e ^{2} W _{0} ( - \frac{2}{e ^{2}} )}{2}

lo g_{5} (x + 2) + lo g_{5} (x - 2) = 1

lo g_{8} (2 n - 3) = lo g_{8} (4 - n)

2 7^{l o g_{9} (x)} = x^{1.5}

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

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

What is the answer to 2x-e^2log_{e}(x)=0 ?
The answer to 2x-e^2log_{e}(x)=0 is x=e^2,x=-(e^2\W_{0}(-2/(e^2)))/2