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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

ln (- 2) + ln (- 5)

ln (\frac{x}{61 \cdot 1 0 ^{- 6}})

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

- lo g_{10} (2.9 \cdot 1 0^{13})

ln (\frac{1.3}{0.04})

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