What is ln(v)-ln|x-1|+2ln|x|+x-1 ?

The solution to ln(v)-ln|x-1|+2ln|x|+x-1 is ln((vx^2)/(|x-1|))+x-1

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simplify ln(v)-ln|x-1|+2ln|x|+x-1

simplify

ln (v) - ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + x - 1

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

Solution steps

ln (v) - ln ∣ x - 1 ∣ + 2 ln ∣ x ∣ + x - 1

Apply log rule:

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

ln (v) - ln ∣ x - 1 ∣ = ln (\frac{v}{∣ x - 1 ∣})

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

Apply log rule:

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

2 ln ∣ x ∣ = ln (∣ x ∣^{2})

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

Apply log rule:

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

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

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

Simplify

\frac{v}{∣ x - 1 ∣} ∣ x ∣^{2} : \frac{v x ^{2}}{∣ x - 1 ∣}

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

ln (ln (x)) - ln (ln (2))

4 ln ∣ x + 5 ∣ - ln ∣ x - 2 ∣ + c

\frac{ln ( 2 ) 32}{5} - \frac{33}{25}

ln (2.05742) + 0.9 \cdot ln (2.036838)

- lo g_{10} (1.14 \cdot 1 0^{- 6})

What is ln(v)-ln|x-1|+2ln|x|+x-1 ?
The solution to ln(v)-ln|x-1|+2ln|x|+x-1 is ln((vx^2)/(|x-1|))+x-1