What is (1/3 ln|e^x+1|-1/3 ln|e^x-2|)*e^{2x} ?

The solution to (1/3 ln|e^x+1|-1/3 ln|e^x-2|)*e^{2x} is ln(\sqrt[3]{(|e^x+1|)/(|e^x-2|)})e^{2x}

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simplify (1/3 ln|e^x+1|-1/3 ln|e^x-2|)*e^{2x}

simplify

(\frac{1}{3} ln ∣ e^{x} + 1 ∣ - \frac{1}{3} ln ∣ e^{x} - 2 ∣) \cdot e^{2 x}

Solution steps

(\frac{1}{3} ln ∣ e^{x} + 1 ∣ - \frac{1}{3} ln ∣ e^{x} - 2 ∣) e^{2 x}

Apply log rule:

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

\frac{1}{3} ln ∣ e^{x} + 1 ∣ = ln (∣ e^{x} + 1 ∣^{\frac{1}{3}})

= e^{2 x} (ln (∣ e^{x} + 1 ∣^{\frac{1}{3}}) - \frac{1}{3} ln ∣ e^{x} - 2 ∣)

Apply log rule:

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

\frac{1}{3} ln ∣ e^{x} - 2 ∣ = ln (∣ e^{x} - 2 ∣^{\frac{1}{3}})

= e^{2 x} (ln (∣ e^{x} + 1 ∣^{\frac{1}{3}}) - ln (∣ e^{x} - 2 ∣^{\frac{1}{3}}))

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (∣ e^{x} + 1 ∣^{\frac{1}{3}}) - ln (∣ e^{x} - 2 ∣^{\frac{1}{3}}) = ln (\frac{∣ e ^{x} + 1 ∣ ^{\frac{1}{3}}}{∣ e ^{x} - 2 ∣ ^{\frac{1}{3}}})

= e^{2 x} ln (\frac{∣ e ^{x} + 1 ∣ ^{\frac{1}{3}}}{∣ e ^{x} - 2 ∣ ^{\frac{1}{3}}})

Combine same powers :

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What is (1/3 ln|e^x+1|-1/3 ln|e^x-2|)*e^{2x} ?
The solution to (1/3 ln|e^x+1|-1/3 ln|e^x-2|)*e^{2x} is ln(\sqrt[3]{(|e^x+1|)/(|e^x-2|)})e^{2x}