What is 2ln(5)-ln(7)+3ln(14)-2ln(10) ?

The solution to 2ln(5)-ln(7)+3ln(14)-2ln(10) is ln(98)

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simplify 2ln(5)-ln(7)+3ln(14)-2ln(10)

simplify

2 ln (5) - ln (7) + 3 ln (14) - 2 ln (10)

ln (98)

Decimal Notation

4.58496 \dots

Solution steps

2 ln (5) - ln (7) + 3 ln (14) - 2 ln (10)

Apply log rule:

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

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

= ln (5^{2}) - ln (7) + 3 ln (14) - 2 ln (10)

Apply log rule:

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

ln (5^{2}) - ln (7) = ln (\frac{5 ^{2}}{7})

= ln (\frac{5 ^{2}}{7}) + 3 ln (14) - 2 ln (10)

5^{2} = 25

= ln (\frac{25}{7}) + 3 ln (14) - 2 ln (10)

Apply log rule:

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

3 ln (14) = ln (1 4^{3})

= ln (\frac{25}{7}) + ln (1 4^{3}) - 2 ln (10)

Apply log rule:

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

ln (\frac{25}{7}) + ln (1 4^{3}) = ln (1 4^{3} \cdot \frac{25}{7})

= ln (1 4^{3} \cdot \frac{25}{7}) - 2 ln (10)

\frac{25}{7} \cdot 1 4^{3} = 9800

= ln (9800) - 2 ln (10)

Apply log rule:

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

2 ln (10) = ln (1 0^{2})

= ln (9800) - ln (1 0^{2})

Apply log rule:

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

ln (9800) - ln (1 0^{2}) = ln (\frac{9800}{1 0 ^{2}})

= ln (\frac{9800}{1 0 ^{2}})

\frac{9800}{1 0 ^{2}} = 98

= ln (98)

lo g_{a} (x) - 2 lo g_{a} (y)

\frac{48125}{( 2 π ( 98 - 94 ) )} ln (\frac{70}{20})

2 lo g_{10} (u) - lo g_{10} (v)

ln (5) - ln (7)

3 ln (x) + 5 ln (z)

What is 2ln(5)-ln(7)+3ln(14)-2ln(10) ?
The solution to 2ln(5)-ln(7)+3ln(14)-2ln(10) is ln(98)