What is ln(u)+3ln(z+1)-2ln(z+2) ?

The solution to ln(u)+3ln(z+1)-2ln(z+2) is ln((u(z+1)^3)/((z+2)^2))

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simplify ln(u)+3ln(z+1)-2ln(z+2)

simplify

ln (u) + 3 ln (z + 1) - 2 ln (z + 2)

ln (\frac{u ( z + 1 ) ^{3}}{( z + 2 ) ^{2}})

Solution steps

ln (u) + 3 ln (z + 1) - 2 ln (z + 2)

Apply log rule:

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

3 ln (z + 1) = ln ((z + 1)^{3})

= ln (u) + ln ((z + 1)^{3}) - 2 ln (z + 2)

Apply log rule:

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

ln (u) + ln ((z + 1)^{3}) = ln (u (z + 1)^{3})

= ln (u (z + 1)^{3}) - 2 ln (z + 2)

Apply log rule:

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

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

= ln (u (z + 1)^{3}) - ln ((z + 2)^{2})

Apply log rule:

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

ln (u (z + 1)^{3}) - ln ((z + 2)^{2}) = ln (\frac{u ( z + 1 ) ^{3}}{( z + 2 ) ^{2}})

= ln (\frac{u ( z + 1 ) ^{3}}{( z + 2 ) ^{2}})

(2 ln (3) - ln (4)) - (2 ln (1) - ln (2))

10 \cdot lo g_{10} (\frac{5 \cdot 1 0 ^{- 6}}{1 \cdot 1 0 ^{- 12}})

100 + 20 \cdot ln (\frac{65}{65 - 23})

ln (\frac{4}{9}) \cdot \frac{20}{ln ( \frac{4}{11} )}

ln ((- 2.772)^{19.764})

What is ln(u)+3ln(z+1)-2ln(z+2) ?
The solution to ln(u)+3ln(z+1)-2ln(z+2) is ln((u(z+1)^3)/((z+2)^2))