What is (ln(4))/(-ln(4)-2ln(3)) ?

The solution to (ln(4))/(-ln(4)-2ln(3)) is -(ln(2))/(ln(6))

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simplify (ln(4))/(-ln(4)-2ln(3))

simplify

\frac{ln ( 4 )}{- ln ( 4 ) - 2 ln ( 3 )}

- \frac{ln ( 2 )}{ln ( 6 )}

Solution steps

\frac{ln ( 4 )}{- ln ( 4 ) - 2 ln ( 3 )}

Apply log rule:

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

2 ln (3) = ln (3^{2})

= \frac{ln ( 4 )}{- ln ( 4 ) - ln ( 3 ^{2} )}

Rewrite

- ln (4) - ln (3^{2})

- (ln (4) + ln (3^{2}))

= \frac{ln ( 4 )}{- ( ln ( 4 ) + ln ( 3 ^{2} ) )}

Apply log rule:

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

ln (4) + ln (3^{2}) = ln (3^{2} \cdot 4)

= \frac{ln ( 4 )}{- ln ( 3 ^{2} \cdot 4 )}

4 \cdot 3^{2} = 36

= \frac{ln ( 4 )}{- ln ( 36 )}

Apply the fraction rule:

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

= - \frac{ln ( 4 )}{ln ( 36 )}

Simplify

ln (36) : 2 ln (6)

= - \frac{ln ( 4 )}{2 ln ( 6 )}

Simplify

ln (4) : 2 ln (2)

= - \frac{2 ln ( 2 )}{2 ln ( 6 )}

Divide the numbers:

\frac{2}{2} = 1

= - \frac{ln ( 2 )}{ln ( 6 )}

ln (a) - 3 ln (b) + 6 ln (c)

ln (xx)

ln (1 n (5 x)^{l n (3 x)})

4 ln (2 e)

ln (x) + ln (y)

What is (ln(4))/(-ln(4)-2ln(3)) ?
The solution to (ln(4))/(-ln(4)-2ln(3)) is -(ln(2))/(ln(6))