What is log_{e}((e^2-2e^{a+2b})/(1-2e^{a+2b)}) ?

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

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simplify log_{e}((e^2-2e^{a+2b})/(1-2e^{a+2b)})

simplify

lo g_{e} (\frac{e ^{2} - 2 e ^{a + 2 b}}{1 - 2 e ^{a + 2 b}})

ln (e^{2} - 2 e^{a + 2 b}) - ln (1 - 2 e^{a + 2 b})

Solution steps

lo g_{e} (\frac{e ^{2} - 2 e ^{a + 2 b}}{1 - 2 e ^{a + 2 b}})

Apply log rule:

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

lo g_{e} (\frac{e ^{2} - 2 e ^{a + 2 b}}{1 - 2 e ^{a + 2 b}}) = lo g_{e} (e^{2} - 2 e^{a + 2 b}) - lo g_{e} (1 - 2 e^{a + 2 b})

= lo g_{e} (e^{2} - 2 e^{a + 2 b}) - lo g_{e} (- 2 e^{a + 2 b} + 1)

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (e^{2} - 2 e^{a + 2 b}) - lo g_{e} (- 2 e^{a + 2 b} + 1)

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (e^{2} - 2 e^{a + 2 b}) - ln (- 2 e^{a + 2 b} + 1)

10 ln (\frac{47}{4})

ln (\frac{g}{2 \cdot v ^{2}})

\frac{80 ln ( 3 ) - 4 ln ( 2 ln ( 3 ) )}{2 ln ( 3 ) - 1}

\frac{6}{5} ln ∣ z - 1 ∣ - \frac{6}{5} ln ∣ z + 4 ∣ + ln ∣ z + 4 ∣ + ln ∣ x ∣

- lo g_{10} (6.64 \cdot 1 0^{- 3})

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