What is ln(3n^2-1)+ln(1/(n^2)) ?

The solution to ln(3n^2-1)+ln(1/(n^2)) is ln((3n^2-1)/(n^2))

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simplify ln(3n^2-1)+ln(1/(n^2))

simplify

ln (3 n^{2} - 1) + ln (\frac{1}{n ^{2}})

ln (\frac{3 n ^{2} - 1}{n ^{2}})

Solution steps

ln (3 n^{2} - 1) + ln (\frac{1}{n ^{2}})

Apply log rule:

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

ln (3 n^{2} - 1) + ln (\frac{1}{n ^{2}}) = ln (\frac{1}{n ^{2}} (3 n^{2} - 1))

= ln (\frac{1}{n ^{2}} (3 n^{2} - 1))

Simplify

(3 n^{2} - 1) \frac{1}{n ^{2}} : \frac{3 n ^{2} - 1}{n ^{2}}

= ln (\frac{3 n ^{2} - 1}{n ^{2}})

4.76 + lo g_{10} (\frac{0.3}{0.2})

15 ln (5) - 24 ln (4) + 9 ln (3) - 2 ln (2)

\frac{1}{10} ln (\frac{1}{3})

ln (5 e^{- \frac{6 x}{t}})

- lo g_{10} (5.4 \cdot 1 0^{- 6})

What is ln(3n^2-1)+ln(1/(n^2)) ?
The solution to ln(3n^2-1)+ln(1/(n^2)) is ln((3n^2-1)/(n^2))