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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (e^{r x}) = r x ln (e)

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

r x ln (e) = r x

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

ln (\frac{2}{2.068})

ln (\frac{s - 4}{s - 3})

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

ln (\frac{40}{100})

ln (z^{2} + 5) - ln (z^{2} - 4)

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