What is (\frac{ln(2)-ln(3))/(ln(3))}{-1} ?

The solution to (\frac{ln(2)-ln(3))/(ln(3))}{-1} is -(ln(2/3))/(ln(3))

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simplify (\frac{ln(2)-ln(3))/(ln(3))}{-1}

simplify

\frac{\frac{l n ( 2 ) - l n ( 3 )}{l n ( 3 )}}{- 1}

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

Decimal Notation

0.36907 \dots

Solution steps

\frac{\frac{l n ( 2 ) - l n ( 3 )}{l n ( 3 )}}{- 1}

Apply log rule:

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

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

= \frac{\frac{l n ( \frac{2}{3} )}{l n ( 3 )}}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{l n ( \frac{2}{3} )}{l n ( 3 )}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{l n ( \frac{2}{3} )}{l n ( 3 )}}{1} = \frac{ln ( \frac{2}{3} )}{ln ( 3 )}

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

lo g_{2 a} (a) + lo g_{3 a} (2 a) + lo g_{4 a} (3 a + 1)

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

ln (\frac{- 1}{0 \cdot 1 - 2})

\frac{6}{5} ln (\frac{y}{x} + 4) - \frac{1}{5} ln (\frac{y}{x} - 1) + ln (x)

\frac{1}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x + 3 ∣ - ln ∣ x ∣

What is (\frac{ln(2)-ln(3))/(ln(3))}{-1} ?
The solution to (\frac{ln(2)-ln(3))/(ln(3))}{-1} is -(ln(2/3))/(ln(3))