What is ln(t-12)-ln(t-11)+ln(2) ?

The solution to ln(t-12)-ln(t-11)+ln(2) is ln((2(t-12))/(t-11))

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simplify ln(t-12)-ln(t-11)+ln(2)

simplify

ln (t - 12) - ln (t - 11) + ln (2)

ln (\frac{2 ( t - 12 )}{t - 11})

Solution steps

ln (t - 12) - ln (t - 11) + ln (2)

Apply log rule:

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

ln (t - 12) - ln (t - 11) = ln (\frac{t - 12}{t - 11})

= ln (\frac{t - 12}{t - 11}) + ln (2)

Apply log rule:

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

ln (\frac{t - 12}{t - 11}) + ln (2) = ln (2 \cdot \frac{t - 12}{t - 11})

= ln (2 \cdot \frac{t - 12}{t - 11})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= ln (\frac{2 ( t - 12 )}{t - 11})

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What is ln(t-12)-ln(t-11)+ln(2) ?
The solution to ln(t-12)-ln(t-11)+ln(2) is ln((2(t-12))/(t-11))