What is log_{10}(15)-2log_{10}(10)+log_{10}(60) ?

The solution to log_{10}(15)-2log_{10}(10)+log_{10}(60) is 2log_{10}(3)

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Popular >

simplify log_{10}(15)-2log_{10}(10)+log_{10}(60)

simplify

lo g_{10} (15) - 2 lo g_{10} (10) + lo g_{10} (60)

2 lo g_{10} (3)

Decimal Notation

0.95424 \dots

Solution steps

lo g_{10} (15) - 2 lo g_{10} (10) + lo g_{10} (60)

Apply log rule:

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

- 2 lo g_{10} (10) = - lo g_{10} (1 0^{2})

= lo g_{10} (15) - lo g_{10} (1 0^{2}) + lo g_{10} (60)

Apply log rule:

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

lo g_{10} (15) - lo g_{10} (1 0^{2}) = lo g_{10} (\frac{15}{1 0 ^{2}})

= lo g_{10} (\frac{15}{1 0 ^{2}}) + lo g_{10} (60)

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

lo g_{10} (\frac{15}{1 0 ^{2}}) + lo g_{10} (60) = lo g_{10} (\frac{15}{1 0 ^{2}} \cdot 60)

= lo g_{10} (\frac{15}{1 0 ^{2}} \cdot 60)

\frac{15}{1 0 ^{2}} \cdot 60 = 9

= lo g_{10} (9)

Factor the number:

9 = 3^{2}

= lo g_{10} (3^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

lo g_{10} (3^{2}) = 2 lo g_{10} (3)

= 2 lo g_{10} (3)

6 lo g_{3} (x) + 12 lo g_{3} (y)

ln (\frac{1}{1 + 2})

lo g_{2} (\frac{a}{b})

6 lo g_{10} (3) - 4 lo g_{10} (2)

What is log_{10}(15)-2log_{10}(10)+log_{10}(60) ?
The solution to log_{10}(15)-2log_{10}(10)+log_{10}(60) is 2log_{10}(3)