What is ln(1+x^2)+1/2 ln(x)-ln(sin(x)) ?

The solution to ln(1+x^2)+1/2 ln(x)-ln(sin(x)) is ln(x^{1/2}csc(x)(1+x^2))

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simplify ln(1+x^2)+1/2 ln(x)-ln(sin(x))

simplify

ln (1 + x^{2}) + \frac{1}{2} ln (x) - ln (sin (x))

ln (x^{\frac{1}{2}} csc (x) (1 + x^{2}))

Solution steps

ln (1 + x^{2}) + \frac{1}{2} ln (x) - ln (sin (x))

Apply log rule:

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

\frac{1}{2} ln (x) = ln (x^{\frac{1}{2}})

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

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

- lo g_{10} (1.74 \cdot 1 0^{- 5})

ln (\frac{( 1 - 2 x )}{1 - x})

- lo g_{10} (3.4 \cdot 1 0^{- 10})

\frac{6}{5} ln ∣ z - 1 ∣ - \frac{1}{5} ln ∣ z + 4 ∣

- lo g_{10} (3.5 \cdot 1 0^{- 3})

What is ln(1+x^2)+1/2 ln(x)-ln(sin(x)) ?
The solution to ln(1+x^2)+1/2 ln(x)-ln(sin(x)) is ln(x^{1/2}csc(x)(1+x^2))