What is ln(y)-1/2 ln(y-1)-1/2 ln(y+1) ?

The solution to ln(y)-1/2 ln(y-1)-1/2 ln(y+1) is ln((ysqrt(y-1)sqrt(y+1))/((y-1)(y+1)))

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simplify ln(y)-1/2 ln(y-1)-1/2 ln(y+1)

simplify

ln (y) - \frac{1}{2} ln (y - 1) - \frac{1}{2} ln (y + 1)

ln (\frac{y y - 1 y + 1}{( y - 1 ) ( y + 1 )})

Solution steps

ln (y) - \frac{1}{2} ln (y - 1) - \frac{1}{2} ln (y + 1)

Apply log rule:

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

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

= ln (y) - ln ((y - 1)^{\frac{1}{2}}) - \frac{1}{2} ln (y + 1)

Apply log rule:

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

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

= ln (\frac{y}{( y - 1 ) ^{\frac{1}{2}}}) - \frac{1}{2} ln (y + 1)

Simplify

\frac{y}{( y - 1 ) ^{\frac{1}{2}}} : \frac{y y - 1}{y - 1}

= ln (\frac{y y - 1}{y - 1}) - \frac{1}{2} ln (y + 1)

Apply log rule:

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

\frac{1}{2} ln (y + 1) = ln ((y + 1)^{\frac{1}{2}})

= ln (\frac{y y - 1}{y - 1}) - ln ((y + 1)^{\frac{1}{2}})

Apply log rule:

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

ln (\frac{y y - 1}{y - 1}) - ln ((y + 1)^{\frac{1}{2}}) = ln (\frac{\frac{y y - 1}{y - 1}}{( y + 1 ) ^{\frac{1}{2}}})

= ln (\frac{\frac{y y - 1}{y - 1}}{( y + 1 ) ^{\frac{1}{2}}})

Simplify

\frac{\frac{y y - 1}{y - 1}}{( y + 1 ) ^{\frac{1}{2}}} : \frac{y y - 1 y + 1}{( y - 1 ) ( y + 1 )}

= ln (\frac{y y - 1 y + 1}{( y - 1 ) ( y + 1 )})

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

- 2 ln ∣ l + d ∣ + ln ∣ d ∣ + ln ∣ 2 l + d ∣

lo g_{4} (x^{2} - 1) - 5 lo g_{4} (x - 1)

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

\frac{10 ln ( 7.01 ) - 10 ln ( 7 )}{7.01 - 7}

What is ln(y)-1/2 ln(y-1)-1/2 ln(y+1) ?
The solution to ln(y)-1/2 ln(y-1)-1/2 ln(y+1) is ln((ysqrt(y-1)sqrt(y+1))/((y-1)(y+1)))