What is the answer to e^{x^x}=2 ?

The answer to e^{x^x}=2 is x=(ln(ln(2)))/(\W_{-1)(ln(ln(2)))},x=(ln(ln(2)))/(\W_{0)(ln(ln(2)))}

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e^{x^x}=2

e^{x^{x}} = 2

x = \frac{ln ( ln ( 2 ) )}{W _{- 1} ( ln ( ln ( 2 ) ) )}, x = \frac{ln ( ln ( 2 ) )}{W _{0} ( ln ( ln ( 2 ) ) )}

Decimal Notation

x = 0.33662 \dots, x = 0.40004 \dots

Solution steps

e^{x^{x}} = 2

Prepare

e^{x^{x}} = 2

for Lambert form:

x e^{- \frac{l n ( l n ( 2 ) )}{x}} = 1

Rewrite the equation with

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

and

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

(\frac{ln ( ln ( 2 ) )}{u}) e^{- u} = 1

Rewrite

(\frac{ln ( ln ( 2 ) )}{u}) e^{- u} = 1

in Lambert form:

u e^{u} = ln (ln (2))

Solve

u e^{u} = ln (ln (2)) : u = W_{- 1} (ln (ln (2))), u = W_{0} (ln (ln (2)))

Verify Solutions:

u = W_{- 1} (ln (ln (2)))

True

, u = W_{0} (ln (ln (2)))

True

The solutions are

u = W_{- 1} (ln (ln (2))), u = W_{0} (ln (ln (2)))

Substitute back

u = \frac{ln ( ln ( 2 ) )}{x},

solve for

x

Solve

\frac{ln ( ln ( 2 ) )}{x} = W_{- 1} (ln (ln (2))) : x = \frac{ln ( ln ( 2 ) )}{W _{- 1} ( ln ( ln ( 2 ) ) )}

Solve

\frac{ln ( ln ( 2 ) )}{x} = W_{0} (ln (ln (2))) : x = \frac{ln ( ln ( 2 ) )}{W _{0} ( ln ( ln ( 2 ) ) )}

x = \frac{ln ( ln ( 2 ) )}{W _{- 1} ( ln ( ln ( 2 ) ) )}, x = \frac{ln ( ln ( 2 ) )}{W _{0} ( ln ( ln ( 2 ) ) )}

3^{x + 1} \cdot 2^{x + 1} = 36

3^{x + 1} = 10

7^{2 x} + 7^{x} - 42 = 0

5^{3 x} = 5^{4 x - 2}

3^{x} = e^{2}

What is the answer to e^{x^x}=2 ?
The answer to e^{x^x}=2 is x=(ln(ln(2)))/(\W_{-1)(ln(ln(2)))},x=(ln(ln(2)))/(\W_{0)(ln(ln(2)))}