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

The solution to (2ln(5)-10ln(9))/(7ln(5)-2ln(9)) is (ln(\frac{25)/(9^{10)})}{ln(78125/81)}

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

simplify

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

\frac{ln ( \frac{25}{9 ^{10}} )}{ln ( \frac{78125}{81} )}

Decimal Notation

- 2.72910 \dots

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

10 ln (9) = ln (9^{10})

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

Apply log rule:

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

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

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

5^{2} = 25

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

Apply log rule:

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

7 ln (5) = ln (5^{7})

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

Apply log rule:

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

2 ln (9) = ln (9^{2})

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

Apply log rule:

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

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

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

\frac{5 ^{7}}{9 ^{2}} = \frac{78125}{81}

= \frac{ln ( \frac{25}{9 ^{10}} )}{ln ( \frac{78125}{81} )}

\frac{ln ( 7 )}{2 ln ( 3 ) - ln ( 7 )}

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

lo g_{14} (\frac{14}{y})

lo g_{10} (\frac{1}{120})

lo g_{10} (\frac{y ^{5}}{x})

What is (2ln(5)-10ln(9))/(7ln(5)-2ln(9)) ?
The solution to (2ln(5)-10ln(9))/(7ln(5)-2ln(9)) is (ln(\frac{25)/(9^{10)})}{ln(78125/81)}