What is 1/3 ln(2x+1)+1/2 [ln(x-4)-ln(x^2+5)] ?

The solution to 1/3 ln(2x+1)+1/2 [ln(x-4)-ln(x^2+5)] is 1/3 ln(2x+1)+1/2 ln((x-4)/(x^2+5))

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simplify 1/3 ln(2x+1)+1/2 [ln(x-4)-ln(x^2+5)]

simplify

\frac{1}{3} ln (2 x + 1) + \frac{1}{2} [ln (x - 4) - ln (x^{2} + 5)]

\frac{1}{3} ln (2 x + 1) + \frac{1}{2} ln (\frac{x - 4}{x ^{2} + 5})

Solution steps

\frac{1}{3} ln (2 x + 1) + \frac{1}{2} [ln (x - 4) - ln (x^{2} + 5)]

Apply log rule:

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

ln (x - 4) - ln (x^{2} + 5) = ln (\frac{x - 4}{x ^{2} + 5})

= \frac{1}{3} ln (2 x + 1) + \frac{1}{2} [ln (\frac{x - 4}{x ^{2} + 5})]

Remove parentheses:

(a) = a

= \frac{1}{3} ln (2 x + 1) + \frac{1}{2} ln (\frac{x - 4}{x ^{2} + 5})

- lo g_{10} (5 (1 0^{- 2}))

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

ln (\frac{1643.85}{821.85}) \cdot 8

ln - \frac{m g}{b} + a - ln (\frac{∣ m ∣ ∣ g ∣}{∣ b ∣})

- lo g_{10} (2.92 \cdot 1 0^{- 14})

What is 1/3 ln(2x+1)+1/2 [ln(x-4)-ln(x^2+5)] ?
The solution to 1/3 ln(2x+1)+1/2 [ln(x-4)-ln(x^2+5)] is 1/3 ln(2x+1)+1/2 ln((x-4)/(x^2+5))