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\def\dd#1#2{\frac{d#1}{d#2}}

\SN{6.4}
\pn1.a) $5^{\log_57}=7$ since $a^{\log_ax}=x$. 
b) $8^{\log_8\sqrt 2}=\sqrt 2$
c) $1.3^{\log_{1.1}75}=75$
d) $\log_4 16=\log_4(4^2)=2$ since $\log_a(a^x)=x$
e) $\log_{3} \sqrt 3=\log_{3} 3^{1/2}=1/2$
f) $\log_4(\frac 14)=\log_4 4^{-1}=-1$.

\pn5.a) $\frac{\log_2x}{\log_3x}=\frac{\frac{\ln x}{\ln2}}{\frac{\ln x}{\ln3}}=
\frac{\ln 3}{\ln2}$.
b) $\frac{\log_2 x}{\log_8 x}=\frac{\frac{\ln x}{\ln2}}{\frac{\ln x}{\ln8}}=
\frac{\ln8}{\ln2}=\frac{\ln2^3}{\ln2}=\frac{3\ln2}{\ln2}=3$
c) $\frac{\log_x a}{\log_{x^2}a}=
\frac{\frac{\ln a}{\ln x}}{\frac{\ln a}{\ln x^2}}=
\frac{\ln x^2}{\ln x}=\frac{2\ln x}{\ln x}=2$.

\pn9. Solve
$3^{\log_3(x^2)}=5e^{\ln x}-3\cdot 10^{\log_{10}(2)}$.
This simplifies to $x^2=5x-3\cdot 2$ or
$x^2-5x+6=0$.  Factoring we see $x=2$ or $x=3$.

\pn13. Find $\dd ys$ where $y=5^{\sqrt{s}}$.
Chain rule: $\dd ys=5^{\sqrt{s}}(\ln 5)\dd{\sqrt s}{s}=
(\ln 5)5^{\sqrt{s}}( \frac 12 s^{-1/2})=
\frac{\ln 5}2 \frac{5^{\sqrt{s}}}{\sqrt s}$.

\pn17. Find $\dd y\theta$ where $y=(\cos \theta)^{\sqrt 2}$.
Power rule:
$\dd y\theta=(\sqrt 2)(\cos \theta)^{\sqrt 2-1}\dd{\cos\theta}{\theta}=$\par
$(\sqrt 2)(\cos \theta)^{\sqrt 2-1}(-\sin\theta)$.

\pn33. Find $\dd yx$ where $y=\log_5{e^x}$.
Chain rule:
$\dd yx=\frac{1}{\ln 5}\frac{1}{e^x}\dd{e^x}x=
\frac{1}{\ln 5}\frac{1}{e^x}e^x=\frac{1}{\ln 5}$.
OR, simplify: $\log_5{e^x}=\frac{\ln e^x}{\ln 5}=
\frac{x}{\ln 5}$ and the derivative is now easy.

\pn39. Find $y^\prime=\dd yx$ if
$y=(x+1)^x$.
Take log of both sides and simplify.
$\ln y=\ln\bigl((x+1)^x\bigr)=x\ln(x+1)$.
Now take the derivative of both sides.
$\frac{y^\prime}{y}=\dd xx \ln(x+1)+x\dd{\ln(x+1)}{x}=
\ln(x+1)+\frac{\dd{x+1}{x}}{x+1}=
\ln(x+1)+\frac 1{x+1}$.
Hence
$y^\prime=(x+1)^x\bigl(\ln(x+1)+\frac x{x+1}\bigr)$.

\pn43. Find $y^\prime=\dd yx$ if
$y=(\sin x)^x$.
Take log of both sides and simplify.
$\ln y=\ln\bigl((\sin x)^x\bigr)=x\ln(\sin x)$.
Now take the derivative of both sides.
$\frac{y^\prime}{y}=\dd xx\ln(\sin x)+x\dd{\ln(\sin x)}x$.
Compute $\dd{\ln(\sin x)}x=\frac{\dd{\sin x}x}{\sin x}=
\frac{-\cos x}{\sin x}=-\cot x$.
Continuing,
$\frac{y^\prime}{y}=\ln(\sin x) -x\cot x$, so
$y^\prime=(\sin x)^x\bigl(\ln(\sin x)-x\cot x\bigr)$.

\pn47. $\int 5^x\> dx=\frac{5^x}{\ln 5} +C$ from formula.

\pn51. $\int_{1}{\sqrt 2} x2^{x^2} dx$.
Substitute $u=x^2$: then $du=2x dx$ so
$\int_{1}{\sqrt 2} x2^{x^2} dx=\int_{1}^{2}\frac 12 2^u du=
\frac 12\bigl(\frac{2^u}{\ln 2}\bigr)\big\vert^2_1=
\frac 12\bigl( \frac{2^2}{\ln 2}-\frac{2^1}{\ln2}\bigr)=
\frac 12\bigl( \frac{4}{\ln 2}-\frac{2}{\ln2}\bigr)=
\frac 12 \frac{2}{\ln 2}=\frac 1{\ln2}$.

\pn57. $\int 3x^{\sqrt 3}dx$.  You knew how to do this one long ago:
$\int 3x^{\sqrt 3}dx=3\frac{x^{\sqrt 3+1}}{\sqrt 3+1}+C$.


\pn65. $\int_0^2 \frac{\log_2(x+2)}{x+2} dx$.
Substitute $u=\log_2(x+2)$ so
$du=\frac1{\ln 2}\frac{\dd{x+2}x}{x+2}dx=\frac1{\ln 2}\frac{1}{x+2}dx$
so $\int_0^2 \frac{\log_2(x+2)}{x+2} dx=
\int_1^2 (\ln 2) u\>du$.
The limits follow since $\log_2(0+2)=1$ and $\log_2(2+2)=\log_2(4)=
\log_2(2^2)=2$.
Continuing, $\int_1^2 (\ln 2) u\>du=
(\ln 2)\frac{u^2}2\Big\vert^2_1=
(\ln 2)\bigl(\frac{2^2}2-\frac{1^2}2\bigr)=
(\ln2)(2-\frac12)=\frac 32\ln2$.


\end
