What is 2log_{3}(2)-3log_{3}(4)+2log_{3}(5) ?

The solution to 2log_{3}(2)-3log_{3}(4)+2log_{3}(5) is log_{3}(25/16)

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simplify 2log_{3}(2)-3log_{3}(4)+2log_{3}(5)

simplify

2 lo g_{3} (2) - 3 lo g_{3} (4) + 2 lo g_{3} (5)

lo g_{3} (\frac{25}{16})

Decimal Notation

0.40622 \dots

Solution steps

2 lo g_{3} (2) - 3 lo g_{3} (4) + 2 lo g_{3} (5)

Apply log rule:

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

3 lo g_{3} (4) = lo g_{3} (4^{3})

= 2 lo g_{3} (2) - lo g_{3} (4^{3}) + 2 lo g_{3} (5)

Apply log rule:

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

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

= 2 lo g_{3} (2) - lo g_{3} (4^{3}) + lo g_{3} (5^{2})

Apply log rule:

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

lo g_{3} (5^{2}) - lo g_{3} (4^{3}) = lo g_{3} (\frac{5 ^{2}}{4 ^{3}})

= 2 lo g_{3} (2) + lo g_{3} (\frac{5 ^{2}}{4 ^{3}})

\frac{5 ^{2}}{4 ^{3}} = \frac{25}{64}

= 2 lo g_{3} (2) + lo g_{3} (\frac{25}{64})

Apply log rule:

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

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

= lo g_{3} (2^{2}) + lo g_{3} (\frac{25}{64})

Apply log rule:

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

lo g_{3} (2^{2}) + lo g_{3} (\frac{25}{64}) = lo g_{3} (2^{2} \cdot \frac{25}{64})

= lo g_{3} (2^{2} \cdot \frac{25}{64})

2^{2} \cdot \frac{25}{64} = \frac{25}{16}

= lo g_{3} (\frac{25}{16})

2 ln (4) + 2 ln (5)

2 lo g_{10} (x - 1) - lo g_{10} (4)

5 + ln (3.6) - 2 ln (4.6)

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

ln (x + 2) - ln (2 x - 1)

What is 2log_{3}(2)-3log_{3}(4)+2log_{3}(5) ?
The solution to 2log_{3}(2)-3log_{3}(4)+2log_{3}(5) is log_{3}(25/16)