What is e^{-(ln(2)+ln(ln(2)))} ?

The solution to e^{-(ln(2)+ln(ln(2)))} is 1/(2ln(2))

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simplify e^{-(ln(2)+ln(ln(2)))}

simplify

e^{- (l n (2) + l n (l n (2)))}

\frac{1}{2 ln ( 2 )}

Decimal Notation

0.72134 \dots

Solution steps

e^{- (l n (2) + l n (l n (2)))}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Remove parentheses:

(a) = a

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

e^{l n (2) + l n (l n (2))} = 2 ln (2)

= \frac{1}{2 ln ( 2 )}

(4.29 x 1 0^{15}) (1.51 x 1 0^{- 4})

e^{- 2 x l n (4 + x^{2})}

\frac{21}{41}

8 (- 3 x^{2} + 1) 0.8 (- 3 x^{2} + 1)

(2.526 \cdot 1 0^{- 5}) \cdot 7

What is e^{-(ln(2)+ln(ln(2)))} ?
The solution to e^{-(ln(2)+ln(ln(2)))} is 1/(2ln(2))