What is 2*log_{10}(2)+log_{10}(0.12) ?

The solution to 2*log_{10}(2)+log_{10}(0.12) is log_{10}(12/25)

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simplify 2*log_{10}(2)+log_{10}(0.12)

simplify

2 \cdot lo g_{10} (2) + lo g_{10} (0.12)

lo g_{10} (\frac{12}{25})

Decimal

- 0.31875 \dots

Solution steps

2 lo g_{10} (2) + lo g_{10} (0.12)

Apply log rule:

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

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

= lo g_{10} (2^{2}) + lo g_{10} (0.12)

Apply log rule:

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

lo g_{10} (2^{2}) + lo g_{10} (0.12) = lo g_{10} (2^{2} \cdot 0.12)

= lo g_{10} (2^{2} \cdot 0.12)

2^{2} \cdot 0.12 = 0.48

= lo g_{10} (0.48)

0.48 = \frac{48}{100}

= lo g_{10} (\frac{48}{100})

Cancel

\frac{48}{100} : \frac{12}{25}

= lo g_{10} (\frac{12}{25})

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What is 2*log_{10}(2)+log_{10}(0.12) ?
The solution to 2*log_{10}(2)+log_{10}(0.12) is log_{10}(12/25)