What is ln((x^{1/2}(2x+7)^4)/((3x^2+1)^2)) ?

The solution to ln((x^{1/2}(2x+7)^4)/((3x^2+1)^2)) is 1/2 ln(x)+4ln(2x+7)-2ln(3x^2+1)

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simplify ln((x^{1/2}(2x+7)^4)/((3x^2+1)^2))

simplify

ln (\frac{x ^{\frac{1}{2}} ( 2 x + 7 ) ^{4}}{( 3 x ^{2} + 1 ) ^{2}})

\frac{1}{2} ln (x) + 4 ln (2 x + 7) - 2 ln (3 x^{2} + 1)

Solution steps

ln (\frac{x ^{\frac{1}{2}} ( 2 x + 7 ) ^{4}}{( 3 x ^{2} + 1 ) ^{2}})

Apply log rule:

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

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

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

Apply log rule:

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

ln ((3 x^{2} + 1)^{2}) = 2 ln (3 x^{2} + 1)

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

Apply log rule:

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

ln (x^{\frac{1}{2}} (2 x + 7)^{4}) = ln (x^{\frac{1}{2}}) + ln ((2 x + 7)^{4})

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

Apply log rule:

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

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

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

Apply log rule:

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

ln ((2 x + 7)^{4}) = 4 ln (2 x + 7)

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

lo g_{10} (1000 x^{3} y^{- 5})

lo g_{2} (3) + 4 lo g_{2} (x) - 6 lo g_{2} (y)

7 lo g_{3} (k) + 6 lo g_{3} (m) - 9 lo g_{3} (n)

ln (x^{2} - 9) - ln (2 x^{2} - 5 x - 3) - 2 ln (x + 3)

- lo g_{10} (2 \cdot 1 0^{- 14})

What is ln((x^{1/2}(2x+7)^4)/((3x^2+1)^2)) ?
The solution to ln((x^{1/2}(2x+7)^4)/((3x^2+1)^2)) is 1/2 ln(x)+4ln(2x+7)-2ln(3x^2+1)