What is (80ln(3)-4ln(2ln(3)))/(2ln(3)-1) ?

The solution to (80ln(3)-4ln(2ln(3)))/(2ln(3)-1) is (ln(\frac{3^{80})/(16ln^4(3)))}{2ln(3)-1}

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simplify (80ln(3)-4ln(2ln(3)))/(2ln(3)-1)

simplify

\frac{80 ln ( 3 ) - 4 ln ( 2 ln ( 3 ) )}{2 ln ( 3 ) - 1}

\frac{ln ( \frac{3 ^{80}}{16 l n ^{4} ( 3 )} )}{2 ln ( 3 ) - 1}

Decimal Notation

70.78054 \dots

Solution steps

\frac{80 ln ( 3 ) - 4 ln ( 2 ln ( 3 ) )}{2 ln ( 3 ) - 1}

Apply log rule:

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

80 ln (3) = ln (3^{80})

= \frac{ln ( 3 ^{80} ) - 4 ln ( 2 ln ( 3 ) )}{2 ln ( 3 ) - 1}

Apply log rule:

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

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

= \frac{ln ( 3 ^{80} ) - ln ( ( 2 ln ( 3 ) ) ^{4} )}{2 ln ( 3 ) - 1}

Apply log rule:

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

ln (3^{80}) - ln ((2 ln (3))^{4}) = ln (\frac{3 ^{80}}{( 2 ln ( 3 ) ) ^{4}})

= \frac{ln ( \frac{3 ^{80}}{( 2 l n ( 3 ) ) ^{4}} )}{2 ln ( 3 ) - 1}

\frac{3 ^{80}}{( 2 ln ( 3 ) ) ^{4}} = \frac{3 ^{80}}{16 ln ^{4} ( 3 )}

= \frac{ln ( \frac{3 ^{80}}{16 l n ^{4} ( 3 )} )}{2 ln ( 3 ) - 1}

\frac{6}{5} ln ∣ z - 1 ∣ - \frac{6}{5} ln ∣ z + 4 ∣ + ln ∣ z + 4 ∣ + ln ∣ x ∣

- lo g_{10} (6.64 \cdot 1 0^{- 3})

5 + 2 ln (2) - 2 ln (5)

\frac{- 3 ln ( 3 ) - 6 ln ( 5 )}{ln ( 5 ) - ln ( 3 )}

ln (z^{2} e^{4 z})

What is (80ln(3)-4ln(2ln(3)))/(2ln(3)-1) ?
The solution to (80ln(3)-4ln(2ln(3)))/(2ln(3)-1) is (ln(\frac{3^{80})/(16ln^4(3)))}{2ln(3)-1}