What is 3log_{0.4}(2)-3log_{0.4}(5) ?

The solution to 3log_{0.4}(2)-3log_{0.4}(5) is log_{0.4}(8/125)

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simplify 3log_{0.4}(2)-3log_{0.4}(5)

simplify

3 lo g_{0.4} (2) - 3 lo g_{0.4} (5)

lo g_{0.4} (\frac{8}{125})

Decimal Notation

3

Solution steps

3 lo g_{0.4} (2) - 3 lo g_{0.4} (5)

Apply log rule:

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

3 lo g_{0.4} (2) = lo g_{0.4} (2^{3})

= lo g_{0.4} (2^{3}) - 3 lo g_{0.4} (5)

Apply log rule:

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

3 lo g_{0.4} (5) = lo g_{0.4} (5^{3})

= lo g_{0.4} (2^{3}) - lo g_{0.4} (5^{3})

Apply log rule:

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

lo g_{0.4} (2^{3}) - lo g_{0.4} (5^{3}) = lo g_{0.4} (\frac{2 ^{3}}{5 ^{3}})

= lo g_{0.4} (\frac{2 ^{3}}{5 ^{3}})

\frac{2 ^{3}}{5 ^{3}} = \frac{8}{125}

= lo g_{0.4} (\frac{8}{125})

3 + 3 ln (\frac{2}{3})

- ln (x) + ln (\frac{( x - 3 ) ^{2}}{x + 1})

- \frac{ln ( 42 )}{ln ( 7 ) - ln ( 6 )}

ln (u) - \frac{1}{2} ln 1 + u^{2} + \frac{1}{2 ( 1 + u ^{2} )} + C

(- \frac{1}{3}) ln (\frac{2}{3})

What is 3log_{0.4}(2)-3log_{0.4}(5) ?
The solution to 3log_{0.4}(2)-3log_{0.4}(5) is log_{0.4}(8/125)