What is (1.0299)*ln(80/250)-(0.41195)ln(550/240) ?

The solution to (1.0299)*ln(80/250)-(0.41195)ln(550/240) is ln(0.21978…)

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simplify (1.0299)*ln(80/250)-(0.41195)ln(550/240)

simplify

(1.0299) \cdot ln (\frac{80}{250}) - (0.41195) ln (\frac{550}{240})

ln (0.21978 \dots)

Decimal Notation

- 1.51512 \dots

Solution steps

(1.0299) ln (\frac{80}{250}) - (0.41195) ln (\frac{550}{240})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

(1.0299) ln (\frac{80}{250}) = ln ((\frac{80}{250})^{(1.0299)})

= ln ((\frac{80}{250})^{(1.0299)}) - (0.41195) ln (\frac{550}{240})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

(0.41195) ln (\frac{550}{240}) = ln ((\frac{550}{240})^{(0.41195)})

= ln ((\frac{80}{250})^{(1.0299)}) - ln ((\frac{550}{240})^{(0.41195)})

Apply log rule:

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

ln ((\frac{80}{250})^{(1.0299)}) - ln ((\frac{550}{240})^{(0.41195)}) = ln (\frac{( \frac{80}{250} ) ^{1.0299}}{( \frac{550}{240} ) ^{0.41195}})

= ln (\frac{( \frac{80}{250} ) ^{1.0299}}{( \frac{550}{240} ) ^{0.41195}})

\frac{( \frac{80}{250} ) ^{(1.0299)}}{( \frac{550}{240} ) ^{(0.41195)}} = 0.21978 \dots

= ln (0.21978 \dots)

ln (\frac{100 + 150}{150})

182000 \cdot [ln (0.5) - ln (1.3)]

ln (\frac{- θ - 1}{θ - 1}) \cdot (- 1)

ln (y - 1) - ln (y + 1)

- ln (\frac{n - i}{i})

What is (1.0299)*ln(80/250)-(0.41195)ln(550/240) ?
The solution to (1.0299)*ln(80/250)-(0.41195)ln(550/240) is ln(0.21978…)