What is |4ln(2)-1|+|4ln(3)-3-4ln(2)| ?

The solution to |4ln(2)-1|+|4ln(3)-3-4ln(2)| is 4ln(2)+2-ln(81/16)

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simplify |4ln(2)-1|+|4ln(3)-3-4ln(2)|

simplify

∣ 4 ln (2) - 1 ∣ + ∣ 4 ln (3) - 3 - 4 ln (2) ∣

4 ln (2) + 2 - ln (\frac{81}{16})

Decimal Notation

3.15072 \dots

Solution steps

∣ 4 ln (2) - 1 ∣ + ∣ 4 ln (3) - 3 - 4 ln (2) ∣

Apply log rule:

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

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

= ∣ 4 ln (2) - 1 ∣ + ln (3^{4}) - 3 - 4 ln (2)

Apply log rule:

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

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

= ∣ 4 ln (2) - 1 ∣ + ln (3^{4}) - 3 - ln (2^{4})

Apply log rule:

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

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

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

\frac{3 ^{4}}{2 ^{4}} = \frac{81}{16}

= ∣ 4 ln (2) - 1 ∣ + ln (\frac{81}{16}) - 3

Apply absolute rule:

∣ A ∣ = A, A \geq 0

∣ 4 ln (2) - 1 ∣ = 4 ln (2) - 1

= 4 ln (2) - 1 + ln (\frac{81}{16}) - 3

Apply absolute rule:

∣ A ∣ = - A, A < 0

ln (\frac{81}{16}) - 3 = - ln (\frac{81}{16}) + 3

= 4 ln (2) - 1 + (- ln (\frac{81}{16}) + 3)

Remove parentheses:

(- a) = - a

= 4 ln (2) - 1 - ln (\frac{81}{16}) + 3

Add/Subtract the numbers:

- 1 + 3 = 2

= 4 ln (2) + 2 - ln (\frac{81}{16})

\frac{2 ln ( \frac{55}{9} )}{ln ( 2 )}

lo g_{10} (\frac{x z ^{2}}{y})

ln (x + 3) - ln (4)

2 ln (10) - ln (5) - 2 ln (2) + ln (3)

- lo g_{10} (8.41 \cdot 0.001)

What is |4ln(2)-1|+|4ln(3)-3-4ln(2)| ?
The solution to |4ln(2)-1|+|4ln(3)-3-4ln(2)| is 4ln(2)+2-ln(81/16)