What is 10ln|P|-10ln|P-1000000| ?

The solution to 10ln|P|-10ln|P-1000000| is ln((P^{10})/((P-1000000)^{10)})

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simplify 10ln|P|-10ln|P-1000000|

simplify

10 ln ∣ P ∣ - 10 ln ∣ P - 1000000 ∣

ln (\frac{P ^{10}}{( P - 1000000 ) ^{10}})

Solution steps

10 ln ∣ P ∣ - 10 ln ∣ P - 1000000 ∣

Apply log rule:

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

10 ln ∣ P ∣ = ln (∣ P ∣^{10})

= ln (∣ P ∣^{10}) - 10 ln ∣ P - 1000000 ∣

Apply log rule:

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

10 ln ∣ P - 1000000 ∣ = ln (∣ P - 1000000 ∣^{10})

= ln (∣ P ∣^{10}) - ln (∣ P - 1000000 ∣^{10})

Apply log rule:

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

ln (∣ P ∣^{10}) - ln (∣ P - 1000000 ∣^{10}) = ln (\frac{∣ P ∣ ^{10}}{∣ P - 1000000 ∣ ^{10}})

= ln (\frac{∣ P ∣ ^{10}}{∣ P - 1000000 ∣ ^{10}})

Simplify

\frac{∣ P ∣ ^{10}}{∣ P - 1000000 ∣ ^{10}} : \frac{P ^{10}}{( P - 1000000 ) ^{10}}

= ln (\frac{P ^{10}}{( P - 1000000 ) ^{10}})

3 lo g_{7} (4) + lo g_{7} (6)

4 lo g_{2} (u) + 2 lo g_{2} (v)

ln (\frac{1}{1 - e ( - ( b + a x + cy ) )})

ln (\frac{3 x + 1 + 9 x ^{2} + 6 x - 8}{3})

ln (\frac{3.64}{23}) + (\frac{6}{23})

What is 10ln|P|-10ln|P-1000000| ?
The solution to 10ln|P|-10ln|P-1000000| is ln((P^{10})/((P-1000000)^{10)})