What is ln(t)egratey^{-3} ?

The solution to ln(t)egratey^{-3} is (e^2atgrln(t))/(y^3)

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simplify ln(t)egratey^{-3}

simplify

ln (t) e g r a t e y^{- 3}

\frac{e ^{2} a t g r ln ( t )}{y ^{3}}

Solution steps

ln (t) e g r a t e y^{- 3}

Apply exponent rule:

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

ee = e^{1 + 1}

= ln (t) g r a t e^{1 + 1} y^{- 3}

Add the numbers:

1 + 1 = 2

= ln (t) g r a t e^{2} y^{- 3}

Apply exponent rule:

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

y^{- 3} = \frac{1}{y ^{3}}

= e^{2} a \frac{1}{y ^{3}} t g r ln (t)

Multiply fractions:

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

= \frac{1 \cdot ln ( t ) g r a t e ^{2}}{y ^{3}}

Multiply:

1 \cdot ln (t) = ln (t)

= \frac{e ^{2} a t g r ln ( t )}{y ^{3}}

2 \cdot 2 + 3

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

6.912 (1 0^{- 4}) \cdot 12

What is ln(t)egratey^{-3} ?
The solution to ln(t)egratey^{-3} is (e^2atgrln(t))/(y^3)