What is 8ln|8x+1|-8ln|x+1| ?

The solution to 8ln|8x+1|-8ln|x+1| is ln(((8x+1)^8)/((x+1)^8))

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simplify 8ln|8x+1|-8ln|x+1|

simplify

8 ln ∣ 8 x + 1 ∣ - 8 ln ∣ x + 1 ∣

ln (\frac{( 8 x + 1 ) ^{8}}{( x + 1 ) ^{8}})

Solution steps

8 ln ∣ 8 x + 1 ∣ - 8 ln ∣ x + 1 ∣

Apply log rule:

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

8 ln ∣ 8 x + 1 ∣ = ln (∣ 8 x + 1 ∣^{8})

= ln (∣ 8 x + 1 ∣^{8}) - 8 ln ∣ x + 1 ∣

Apply log rule:

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

8 ln ∣ x + 1 ∣ = ln (∣ x + 1 ∣^{8})

= ln (∣ 8 x + 1 ∣^{8}) - ln (∣ x + 1 ∣^{8})

Apply log rule:

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

ln (∣ 8 x + 1 ∣^{8}) - ln (∣ x + 1 ∣^{8}) = ln (\frac{∣ 8 x + 1 ∣ ^{8}}{∣ x + 1 ∣ ^{8}})

= ln (\frac{∣ 8 x + 1 ∣ ^{8}}{∣ x + 1 ∣ ^{8}})

Simplify

\frac{∣ 8 x + 1 ∣ ^{8}}{∣ x + 1 ∣ ^{8}} : \frac{( 8 x + 1 ) ^{8}}{( x + 1 ) ^{8}}

= ln (\frac{( 8 x + 1 ) ^{8}}{( x + 1 ) ^{8}})

ln (\frac{k}{n}) - e^{a t + c}

ln (x (x - 3)^{2})

ln (x) - ln (y) - ln (z)

lo g_{9} (2 x + 3) - lo g_{9} (\frac{1}{2 x - 3})

What is 8ln|8x+1|-8ln|x+1| ?
The solution to 8ln|8x+1|-8ln|x+1| is ln(((8x+1)^8)/((x+1)^8))