What is 9log_{n}(c)-log_{n}(q) ?

The solution to 9log_{n}(c)-log_{n}(q) is log_{n}((c^9)/q)

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simplify 9log_{n}(c)-log_{n}(q)

simplify

9 lo g_{n} (c) - lo g_{n} (q)

lo g_{n} (\frac{c ^{9}}{q})

Solution steps

9 lo g_{n} (c) - lo g_{n} (q)

Apply log rule:

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

9 lo g_{n} (c) = lo g_{n} (c^{9})

= lo g_{n} (c^{9}) - lo g_{n} (q)

Apply log rule:

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

lo g_{n} (c^{9}) - lo g_{n} (q) = lo g_{n} (\frac{c ^{9}}{q})

= lo g_{n} (\frac{c ^{9}}{q})

ln (15 \cdot e^{\frac{x}{T}})

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

- \frac{5}{9} (ln (10 - 0.9 v^{2}) - ln (10 - 0.9 v_{0}^{2}))

\frac{1}{2} lo g_{10} (11)

ln (\frac{( x + 5 ) ^{2}}{( x - 9 ) ^{2} ( x ^{2} - 15 ) ^{5}})

What is 9log_{n}(c)-log_{n}(q) ?
The solution to 9log_{n}(c)-log_{n}(q) is log_{n}((c^9)/q)