Bern's logarithm

$\begin{aligned} s & = \int_3^5 \log_{16-(x-4)^2}(x^3) \ dx & \small \color{#3D99F6} \text{Note that }\log_a x = \frac {\log x}{\log a} \\ & = \int_3^5 \frac {3\ln x}{\ln (16-(x^2-8x+16))} \ dx \\ & = 3\int_3^5 \frac {\ln x}{\ln (8x-x^2)} \ dx & \small \color{#3D99F6} \text{Note that }\log_ab = \log a + \log b \\ & = 3\int_3^5 \frac {\ln x}{\ln x + \ln(8-x)} \ dx & \small \color{#3D99F6} \text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 32 \int_3^5 \left(\frac {\ln x}{\ln x + \ln(8-x)} + \frac {\ln (8-x)}{\ln (8-x) + \ln x} \right) dx \\ & = \frac 32 \int_3^5 \ dx = \frac 32 x \ \bigg|_3^5 = \boxed{3} \end{aligned}$

$\large \displaystyle s = \int_{3}^{5} \log_{16 - (x-4)^2} (x^3) dx$

• Since $\color{#20A900} \displaystyle \log_a b = \frac{\ln(b)}{\ln(a)}$ :

$\large \displaystyle s = \int_{3}^{5} \frac{\ln(x^3)}{\ln(8x - x^2)}dx$

$\large \displaystyle s = 3 \int_{3}^{5} \frac{\ln(x)}{\ln(8 - x) + \ln(x)}dx$

• Since $\color{#20A900} \displaystyle \int_a^b f(x) = \int_a^b f(a+b-x)$ :

$\large \displaystyle s = 3 \cdot \frac12 \left [ \int_{3}^{5} \frac{\ln(x)}{\ln(8-x) + \ln(x)}dx + \int_{3}^{5} \frac{\ln(8-x)}{\ln(x) + \ln(8-x)}dx \right ]$

$\large \displaystyle s = 3 \cdot \frac12 \left [ \int_{3}^{5} dx \right ]$

$\color{#3D99F6} \large \displaystyle s = 3, \boxed {\large \displaystyle 7s = 21}$