What is (0-ln(2))-(2ln(2)-ln(3)) ?

The solution to (0-ln(2))-(2ln(2)-ln(3)) is -ln(8/3)

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simplify (0-ln(2))-(2ln(2)-ln(3))

simplify

(0 - ln (2)) - (2 ln (2) - ln (3))

- ln (\frac{8}{3})

Decimal Notation

- 0.98082 \dots

Solution steps

(0 - ln (2)) - (2 ln (2) - ln (3))

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln (2) = ln (2^{2})

= (- ln (2) + 0) - (ln (2^{2}) - ln (3))

Apply log rule:

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

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

= 0 - ln (2) - ln (\frac{2 ^{2}}{3})

2^{2} = 4

= 0 - ln (2) - ln (\frac{4}{3})

Rewrite

- ln (2) - ln (\frac{4}{3})

- (ln (2) + ln (\frac{4}{3}))

= 0 - (ln (2) + ln (\frac{4}{3}))

Apply log rule:

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

ln (2) + ln (\frac{4}{3}) = ln (2 \cdot \frac{4}{3})

= 0 - ln (2 \cdot \frac{4}{3})

2 \cdot \frac{4}{3} = \frac{8}{3}

= 0 - ln (\frac{8}{3})

0 - ln (\frac{8}{3}) = - ln (\frac{8}{3})

= - ln (\frac{8}{3})

- 20 ln (\frac{33}{7})

2 ln (1 + 2)

- \frac{1}{5} ln ∣ y + 1 ∣ + \frac{1}{5} ln ∣ y - 4 ∣

- lo g_{10} (6.35 \cdot 1 0^{- 8})

3 \cdot lo g_{e} (10 x)

What is (0-ln(2))-(2ln(2)-ln(3)) ?
The solution to (0-ln(2))-(2ln(2)-ln(3)) is -ln(8/3)