What is 2ln(5)-3ln(3)+ln(5)-ln(3) ?

The solution to 2ln(5)-3ln(3)+ln(5)-ln(3) is ln(125/81)

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

simplify

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

ln (\frac{125}{81})

Decimal Notation

0.43386 \dots

Solution steps

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

Apply log rule:

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

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

= ln (5^{2}) - 3 ln (3) + ln (5) - ln (3)

Apply log rule:

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

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

= ln (5^{2}) - ln (3^{3}) + ln (5) - ln (3)

Apply log rule:

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

ln (5^{2}) - ln (3^{3}) = ln (\frac{5 ^{2}}{3 ^{3}})

= ln (\frac{5 ^{2}}{3 ^{3}}) + ln (5) - ln (3)

\frac{5 ^{2}}{3 ^{3}} = \frac{25}{27}

= ln (\frac{25}{27}) + ln (5) - ln (3)

Apply log rule:

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

ln (\frac{25}{27}) + ln (5) = ln (5 \cdot \frac{25}{27})

= ln (5 \cdot \frac{25}{27}) - ln (3)

\frac{25}{27} \cdot 5 = \frac{125}{27}

= ln (\frac{125}{27}) - ln (3)

Apply log rule:

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

ln (\frac{125}{27}) - ln (3) = ln (\frac{\frac{125}{27}}{3})

= ln (\frac{\frac{125}{27}}{3})

\frac{\frac{125}{27}}{3} = \frac{125}{81}

= ln (\frac{125}{81})

ln (\frac{1 + 4}{2})

ln (2) + 5 ln (x)

ln (e) + ln (2 x) - ln (3 y) - 2 ln (z)

ln (z \frac{y}{z})

ln (b) - ln (\frac{z}{a})

What is 2ln(5)-3ln(3)+ln(5)-ln(3) ?
The solution to 2ln(5)-3ln(3)+ln(5)-ln(3) is ln(125/81)