What is e^{-2ln(ln(x))-1ln(x)+c} ?

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

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simplify e^{-2ln(ln(x))-1ln(x)+c}

simplify

e^{- 2 l n (l n (x)) - 1 l n (x) + c}

\frac{e ^{c}}{x ln ^{2} ( x )}

Solution steps

e^{- 2 l n (l n (x)) - 1 \cdot l n (x) + c}

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

= e^{- l n (l n^{2} (x)) - 1 \cdot l n (x) + c}

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

1 \cdot ln (x) = ln (x^{1})

= e^{- l n (l n^{2} (x)) - l n (x^{1}) + c}

Rewrite

- ln (ln^{2} (x)) - ln (x^{1})

- (ln (ln^{2} (x)) + ln (x^{1}))

= e^{- (l n (l n^{2} (x)) + l n (x^{1})) + c}

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln (ln^{2} (x)) + ln (x^{1}) = ln (x^{1} ln^{2} (x))

= e^{- l n (x^{1} l n^{2} (x)) + c}

Apply rule

a^{1} = a

x^{1} = x

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

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

= e^{- l n (x l n^{2} (x))} e^{c}

Simplify

e^{- l n (x l n^{2} (x))} : (x ln^{2} (x))^{- 1}

= (x ln^{2} (x))^{- 1} e^{c}

Apply exponent rule:

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

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

= e^{c} \frac{1}{x ln ^{2} ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot e ^{c}}{x ln ^{2} ( x )}

Multiply:

1 \cdot e^{c} = e^{c}

= \frac{e ^{c}}{x ln ^{2} ( x )}

- lo g_{10} (2.51 \cdot 1 0^{- 7})

2 lo g_{10} (x - 3) - lo g_{10} (x + 1)

ln (x 50 0^{- x})

5 lo g_{10} (x) - 4 lo g_{10} (x + 8)

ln (c^{1}) + 0.8 \cdot ln (c^{2})

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