What is simplify-ln(2)+ln(11)-ln(10)-ln(55) ?

The solution to simplify-ln(2)+ln(11)-ln(10)-ln(55) is -2ln(10)

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simplify-ln(2)+ln(11)-ln(10)-ln(55)

simplify

- ln (2) + ln (11) - ln (10) - ln (55)

- 2 ln (10)

Decimal Notation

- 4.60517 \dots

Solution steps

- ln (2) + ln (11) - ln (10) - ln (55)

Apply log rule:

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

ln (11) - ln (2) = ln (\frac{11}{2})

= ln (\frac{11}{2}) - ln (10) - ln (55)

Apply log rule:

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

ln (\frac{11}{2}) - ln (10) = ln (\frac{\frac{11}{2}}{10})

= ln (\frac{\frac{11}{2}}{10}) - ln (55)

\frac{\frac{11}{2}}{10} = \frac{11}{20}

= ln (\frac{11}{20}) - ln (55)

Apply log rule:

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

ln (\frac{11}{20}) - ln (55) = ln (\frac{\frac{11}{20}}{55})

= ln (\frac{\frac{11}{20}}{55})

\frac{\frac{11}{20}}{55} = \frac{1}{100}

= ln (\frac{1}{100})

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (100)

Rewrite

100

in power-base form:

100 = 1 0^{2}

= - ln (1 0^{2})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

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

= - 2 ln (10)

ln (\frac{3.64}{23}) + \frac{3}{23} ln (23)

\frac{1}{3} ln (\frac{17}{3} + 35)

\frac{1}{3} (ln (x) - ln (y))

ln (\frac{2.5}{3})

lo g_{4} (\frac{x}{16})

What is simplify-ln(2)+ln(11)-ln(10)-ln(55) ?
The solution to simplify-ln(2)+ln(11)-ln(10)-ln(55) is -2ln(10)