What is (1/x)^2+log_{e}(2x) ?

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

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simplify (1/x)^2+log_{e}(2x)

simplify

(\frac{1}{x})^{2} + lo g_{e} (2 x)

\frac{1}{x ^{2}} + ln (2) + ln (x)

Solution steps

(\frac{1}{x})^{2} + lo g_{e} (2 x)

Apply log rule:

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

lo g_{e} (2 x) = lo g_{e} (2) + lo g_{e} (x)

= (\frac{1}{x})^{2} + lo g_{e} (2) + lo g_{e} (x)

(\frac{1}{x})^{2} = \frac{1}{x ^{2}}

lo g_{e} (2) = ln (2)

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

= \frac{1}{x ^{2}} + ln (2) + ln (x)

ln (\frac{1}{2 e ^{x^{2}}})

ln (3 x + 16) + ln (x + 4)

- 4 ln ∣ x ∣ - 3 ln ∣ 2 x + 1 ∣ + 11 ln ∣ x + 1 ∣

2 lo g_{2} (x) + 8 lo g_{2} (y)

- ln (\frac{x + 1}{x - 1})

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