What is log_{1.12}(125/12)+1 ?

The solution to log_{1.12}(125/12)+1 is 3log_{1.12}(5)-log_{1.12}(12)+1

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simplify log_{1.12}(125/12)+1

simplify

lo g_{1.12} (\frac{125}{12}) + 1

3 lo g_{1.12} (5) - lo g_{1.12} (12) + 1

Decimal Notation

21.67796 \dots

Solution steps

lo g_{1.12} (\frac{125}{12}) + 1

Apply log rule:

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

lo g_{1.12} (\frac{125}{12}) = lo g_{1.12} (125) - lo g_{1.12} (12)

= lo g_{1.12} (125) - lo g_{1.12} (12) + 1

lo g_{1.12} (125) = 3 lo g_{1.12} (5)

= 3 lo g_{1.12} (5) - lo g_{1.12} (12) + 1

0.3 \cdot ln (3400) + 0.7 \cdot ln (5800)

40 e^{\frac{l n ( 53 ) 31}{21}}

ln (\frac{1}{x ^{2} + 9})

(lo g_{30} (2) + lo g_{30} (3) + lo g_{2} (12))^{2}

ln (\frac{x ^{2} - 25}{x ^{4}})

What is log_{1.12}(125/12)+1 ?
The solution to log_{1.12}(125/12)+1 is 3log_{1.12}(5)-log_{1.12}(12)+1