What is ln(2sqrt(8))0.125+ln(3/5)0.6-49ln(7e^3) ?

The solution to ln(2sqrt(8))0.125+ln(3/5)0.6-49ln(7e^3) is ln(5.12708E-106)

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simplify ln(2sqrt(8))0.125+ln(3/5)0.6-49ln(7e^3)

simplify

ln (28) 0.125 + ln (\frac{3}{5}) 0.6 - 49 ln (7 e^{3})

ln (5.12708 E - 106)

Decimal Notation

- 242.43948 \dots

Solution steps

ln (28) \cdot 0.125 + ln (\frac{3}{5}) \cdot 0.6 - 49 ln (7 e^{3})

Apply log rule:

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

0.125 ln (28) = ln ((28)^{0.125})

= ln ((28)^{0.125}) + 0.6 ln (\frac{3}{5}) - 49 ln (7 e^{3})

Apply log rule:

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

0.6 ln (\frac{3}{5}) = ln ((\frac{3}{5})^{0.6})

= ln ((28)^{0.125}) + ln ((\frac{3}{5})^{0.6}) - 49 ln (7 e^{3})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln ((28)^{0.125}) + ln ((\frac{3}{5})^{0.6}) = ln ((\frac{3}{5})^{0.6} (28)^{0.125})

= ln ((\frac{3}{5})^{0.6} (28)^{0.125}) - 49 ln (7 e^{3})

(28)^{0.125} (\frac{3}{5})^{0.6} = 0.91403 \dots

= ln (0.91403 \dots) - 49 ln (7 e^{3})

Apply log rule:

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

49 ln (7 e^{3}) = ln ((7 e^{3})^{49})

= ln (0.91403 \dots) - ln ((7 e^{3})^{49})

Apply log rule:

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

ln (0.91403 \dots) - ln ((7 e^{3})^{49}) = ln (\frac{0.91403 \dots}{( 7 e ^{3} ) ^{49}})

= ln (\frac{0.91403 \dots}{( 7 e ^{3} ) ^{49}})

\frac{0.91403 \dots}{( 7 e ^{3} ) ^{49}} = 5.12708 E - 106

= ln (5.12708 E - 106)

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lo g_{10} (\frac{10}{a ^{2} + b ^{2}})

5 lo g_{2} (x) + \frac{3}{4} lo g_{2} (y) - lo g_{2} (z)

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What is ln(2sqrt(8))0.125+ln(3/5)0.6-49ln(7e^3) ?
The solution to ln(2sqrt(8))0.125+ln(3/5)0.6-49ln(7e^3) is ln(5.12708E-106)