What is [5log_{7}(2)+3log_{7}(m-2)] ?

The solution to [5log_{7}(2)+3log_{7}(m-2)] is log_{7}(32(m-2)^3)

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simplify [5log_{7}(2)+3log_{7}(m-2)]

simplify

[5 lo g_{7} (2) + 3 lo g_{7} (m - 2)]

lo g_{7} (32 (m - 2)^{3})

Solution steps

[5 lo g_{7} (2) + 3 lo g_{7} (m - 2)]

Apply log rule:

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

5 lo g_{7} (2) = lo g_{7} (2^{5})

= [lo g_{7} (2^{5}) + 3 lo g_{7} (m - 2)]

Apply log rule:

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

3 lo g_{7} (m - 2) = lo g_{7} ((m - 2)^{3})

= [lo g_{7} (2^{5}) + lo g_{7} ((m - 2)^{3})]

Apply log rule:

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

lo g_{7} (2^{5}) + lo g_{7} ((m - 2)^{3}) = lo g_{7} (2^{5} (m - 2)^{3})

= [lo g_{7} (2^{5} (m - 2)^{3})]

2^{5} = 32

= [lo g_{7} (32 (m - 2)^{3})]

Remove parentheses:

(a) = a

= lo g_{7} (32 (m - 2)^{3})

lo g_{4} (1 e^{20})

\frac{d}{d x} (\frac{1}{x ^{2} + 1}) + ln (\frac{1}{x})

\frac{7}{30} (ln (7) - ln (2))

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

ln (2) e^{- 6} + (- 2.268) \cdot ln (0.15)

What is [5log_{7}(2)+3log_{7}(m-2)] ?
The solution to [5log_{7}(2)+3log_{7}(m-2)] is log_{7}(32(m-2)^3)