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

The solution to 2log_{3}(2)-3log_{3}(4)-2log_{3}(5) is -2log_{3}(20)

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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)

- 2 lo g_{3} (20)

Decimal Notation

- 5.45366 \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})

Rewrite

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

- (lo g_{3} (4^{3}) + 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} (ab)

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

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

4^{3} \cdot 5^{2} = 1600

= 2 lo g_{3} (2) - lo g_{3} (1600)

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} (1600)

Apply log rule:

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

lo g_{3} (2^{2}) - lo g_{3} (1600) = lo g_{3} (\frac{2 ^{2}}{1600})

= lo g_{3} (\frac{2 ^{2}}{1600})

\frac{2 ^{2}}{1600} = \frac{1}{400}

= lo g_{3} (\frac{1}{400})

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - lo g_{3} (400)

Rewrite

400

in power-base form:

400 = 2 0^{2}

= - lo g_{3} (2 0^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

lo g_{3} (2 0^{2}) = 2 lo g_{3} (20)

= - 2 lo g_{3} (20)

ln (19) - ln (44)

ln (0.5833) - ln (1.625)

ln (x) + 9 ln (y)

\frac{1}{10} \cdot ln (y) - \frac{1}{10} \cdot ln (10 - y)

lo g_{7} ((x y)^{5})

What is 2log_{3}(2)-3log_{3}(4)-2log_{3}(5) ?
The solution to 2log_{3}(2)-3log_{3}(4)-2log_{3}(5) is -2log_{3}(20)