What is log_{10}(11.1+3)+log_{10}(11.1-4) ?

The solution to log_{10}(11.1+3)+log_{10}(11.1-4) is log_{10}(10011/100)

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simplify log_{10}(11.1+3)+log_{10}(11.1-4)

simplify

lo g_{10} (11.1 + 3) + lo g_{10} (11.1 - 4)

lo g_{10} (\frac{10011}{100})

Decimal Notation

2.00047 \dots

Solution steps

lo g_{10} (11.1 + 3) + lo g_{10} (11.1 - 4)

Apply log rule:

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

lo g_{10} (11.1 + 3) + lo g_{10} (11.1 - 4) = lo g_{10} ((11.1 + 3) (11.1 - 4))

= lo g_{10} ((11.1 + 3) (11.1 - 4))

(11.1 + 3) (11.1 - 4) = 100.11

= lo g_{10} (100.11)

100.11 = \frac{10011}{100}

= lo g_{10} (\frac{10011}{100})

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\frac{( 6 ln ( x ) + 16 - 6 ln ( 4 ) )}{x}

- ln (0.8) + 2 ln (0.75) - 8 ln (0.65) + ln (0.6)

What is log_{10}(11.1+3)+log_{10}(11.1-4) ?
The solution to log_{10}(11.1+3)+log_{10}(11.1-4) is log_{10}(10011/100)