What is ln(a)-3ln(b)+5ln(c) ?

The solution to ln(a)-3ln(b)+5ln(c) is ln((ac^5)/(b^3))

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simplify ln(a)-3ln(b)+5ln(c)

simplify

ln (a) - 3 ln (b) + 5 ln (c)

ln (\frac{a c ^{5}}{b ^{3}})

Solution steps

ln (a) - 3 ln (b) + 5 ln (c)

Apply log rule:

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

3 ln (b) = ln (b^{3})

= ln (a) - ln (b^{3}) + 5 ln (c)

Apply log rule:

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

ln (a) - ln (b^{3}) = ln (\frac{a}{b ^{3}})

= ln (\frac{a}{b ^{3}}) + 5 ln (c)

Apply log rule:

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

5 ln (c) = ln (c^{5})

= ln (\frac{a}{b ^{3}}) + ln (c^{5})

Apply log rule:

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

ln (\frac{a}{b ^{3}}) + ln (c^{5}) = ln (c^{5} \frac{a}{b ^{3}})

= ln (c^{5} \frac{a}{b ^{3}})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= ln (\frac{a c ^{5}}{b ^{3}})

ln (- R) + ln (R 2)

- lo g_{10} (1 \cdot 1 0^{- 0.9})

ln (0.02 (2000))

ln (\frac{( e ^{- u} u ^{x} )}{x !})

lo g_{4} (\frac{1}{126})

What is ln(a)-3ln(b)+5ln(c) ?
The solution to ln(a)-3ln(b)+5ln(c) is ln((ac^5)/(b^3))