What is 5ln(2)-2ln(7/2)+9/2-25ln(10/7) ?

The solution to 5ln(2)-2ln(7/2)+9/2-25ln(10/7) is ln((7^{23})/(5^{25)*262144})+9/2

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simplify 5ln(2)-2ln(7/2)+9/2-25ln(10/7)

simplify

5 ln (2) - 2 ln (\frac{7}{2}) + \frac{9}{2} - 25 ln (\frac{10}{7})

ln (\frac{7 ^{23}}{5 ^{25} \cdot 262144}) + \frac{9}{2}

Decimal Notation

- 3.45666 \dots

Solution steps

5 ln (2) - 2 ln (\frac{7}{2}) + \frac{9}{2} - 25 ln (\frac{10}{7})

Apply log rule:

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

5 ln (2) = ln (2^{5})

= ln (2^{5}) - 2 ln (\frac{7}{2}) + \frac{9}{2} - 25 ln (\frac{10}{7})

Apply log rule:

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

2 ln (\frac{7}{2}) = ln ((\frac{7}{2})^{2})

= ln (2^{5}) - ln ((\frac{7}{2})^{2}) + \frac{9}{2} - 25 ln (\frac{10}{7})

Apply log rule:

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

ln (2^{5}) - ln ((\frac{7}{2})^{2}) = ln (\frac{2 ^{5}}{( \frac{7}{2} ) ^{2}})

= ln (\frac{2 ^{5}}{( \frac{7}{2} ) ^{2}}) + \frac{9}{2} - 25 ln (\frac{10}{7})

\frac{2 ^{5}}{( \frac{7}{2} ) ^{2}} = \frac{128}{49}

= ln (\frac{128}{49}) + \frac{9}{2} - 25 ln (\frac{10}{7})

Apply log rule:

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

25 ln (\frac{10}{7}) = ln ((\frac{10}{7})^{25})

= ln (\frac{128}{49}) + \frac{9}{2} - ln ((\frac{10}{7})^{25})

Apply log rule:

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

ln (\frac{128}{49}) - ln ((\frac{10}{7})^{25}) = ln (\frac{\frac{128}{49}}{( \frac{10}{7} ) ^{25}})

= ln (\frac{\frac{128}{49}}{( \frac{10}{7} ) ^{25}}) + \frac{9}{2}

\frac{\frac{128}{49}}{( \frac{10}{7} ) ^{25}} = \frac{7 ^{23}}{5 ^{25} \cdot 262144}

= ln (\frac{7 ^{23}}{5 ^{25} \cdot 262144}) + \frac{9}{2}

ln^{2} (\frac{1}{3})

4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 3 + \frac{1}{6 ( x ^{2} + 3 )})

- lo g_{10} (2.6 \cdot 1 0^{- 7})

ln (d) - ln (a \cdot b) + ln (c)

6 lo g_{7} (7 x + 1) + \frac{1}{4} lo g_{7} (x + 4)

What is 5ln(2)-2ln(7/2)+9/2-25ln(10/7) ?
The solution to 5ln(2)-2ln(7/2)+9/2-25ln(10/7) is ln((7^{23})/(5^{25)*262144})+9/2