What is 1/3 ln|f^7(x)|+29/21 ln|f(x)+15|+C ?

The solution to 1/3 ln|f^7(x)|+29/21 ln|f(x)+15|+C is ln(|fx+15|^{29/21}(|x||f^7|)^{1/3})+C

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simplify 1/3 ln|f^7(x)|+29/21 ln|f(x)+15|+C

simplify

\frac{1}{3} ln f^{7} (x) + \frac{29}{21} ln ∣ f (x) + 15 ∣ + C

ln (∣ f x + 15 ∣^{\frac{29}{21}} (∣ x ∣ f^{7})^{\frac{1}{3}}) + C

Solution steps

\frac{1}{3} ln f^{7} (x) + \frac{29}{21} ln ∣ f (x) + 15 ∣ + C

Apply log rule:

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

\frac{1}{3} ln f^{7} (x) = ln (f^{7} (x)^{\frac{1}{3}})

= ln (f^{7} (x)^{\frac{1}{3}}) + \frac{29}{21} ln ∣ f (x) + 15 ∣ + C

Apply log rule:

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

\frac{29}{21} ln ∣ f (x) + 15 ∣ = ln (∣ f (x) + 15 ∣^{\frac{29}{21}})

= ln (f^{7} (x)^{\frac{1}{3}}) + ln (∣ f (x) + 15 ∣^{\frac{29}{21}}) + C

Apply log rule:

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

ln (f^{7} (x)^{\frac{1}{3}}) + ln (∣ f (x) + 15 ∣^{\frac{29}{21}}) = ln (∣ f x + 15 ∣^{\frac{29}{21}} f^{7} x^{\frac{1}{3}})

= ln (∣ f x + 15 ∣^{\frac{29}{21}} f^{7} x^{\frac{1}{3}}) + C

Simplify

f^{7} (x)^{\frac{1}{3}} ∣ f (x) + 15 ∣^{\frac{29}{21}} : ∣ f x + 15 ∣^{\frac{29}{21}} (∣ x ∣ f^{7})^{\frac{1}{3}}

= ln (∣ f x + 15 ∣^{\frac{29}{21}} (∣ x ∣ f^{7})^{\frac{1}{3}}) + C

0.6 \cdot lo g_{2} (0.6) + 0.4 \cdot lo g_{2} (0.4)

ln (3) + 5 ln (x) - ln (y)

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

0.5 (- \frac{3}{5} lo g_{2} (\frac{3}{5}) - \frac{2}{5} lo g_{2} (\frac{2}{5}))

(\frac{( ln ( x ) - ln ( 100 ) )}{2 \cdot ln ( 2 )})^{2}

What is 1/3 ln|f^7(x)|+29/21 ln|f(x)+15|+C ?
The solution to 1/3 ln|f^7(x)|+29/21 ln|f(x)+15|+C is ln(|fx+15|^{29/21}(|x||f^7|)^{1/3})+C