What is log_{10}(i*(e^2-1)) ?

The solution to log_{10}(i*(e^2-1)) is log_{10}(i)+log_{10}(e^2-1)

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simplify log_{10}(i*(e^2-1))

simplify

lo g_{10} (i \cdot (e^{2} - 1))

lo g_{10} (i) + lo g_{10} (e^{2} - 1)

Solution steps

lo g_{10} (i (e^{2} - 1))

Apply log rule:

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

lo g_{10} (i (e^{2} - 1)) = lo g_{10} (i) + lo g_{10} (e^{2} - 1)

= lo g_{10} (i) + lo g_{10} (e^{2} - 1)

2 + lo g_{10} (\frac{5.25}{5.5})

ln (x + 3 sin (x))

36 lo g_{3} (x) - 6 lo g_{3} (y)

2 ln (x) + ln (x - 4) - ln (y)

ln (\frac{40}{6})

What is log_{10}(i*(e^2-1)) ?
The solution to log_{10}(i*(e^2-1)) is log_{10}(i)+log_{10}(e^2-1)