Partial Solutions

Since \( y=\log_c (x) \) is equivalent to \( c^y=x \), we have:

1. \( \log_6 (216)=3 \) equivalent to \( 6^3=216 \)
2. \( \log_{10} (1000)=3 \) equivalent to \( 10^3=1000 \)
3. \( \log_{25} \left(\frac{1}{625}\right)=-2 \) equivalent to \( 25^{-2}=\frac{1}{625} \)
4. \( \log_{16} \left(4\sqrt{2}\right)=\frac{5}{8} \) equivalent to \( 16^{\frac{5}{8}}=4\sqrt{2} \)

1. We must find \( a \) such that \( 3^a=35 \). Since \( 3^3\lt35 \lt 3^4 \) and \( 35 \) is closer in value to \( 27 \) than \( 81 \), we check,\[\begin{align*} 3^{3.2} &\approx 33.6347 \\ 3^{3.3} &\approx 37.5401 \\ 3^{3.25} &\approx 35.5340 \\ 3^{3.24} &\approx 35.1458 \\ 3^{3.23} &\approx 34.7618 \end{align*}\] Therefore, \( \log_3 (35)\approx 3.24 \) to two decimal places of accuracy.
2. We must find \( a \) such that \( 2^a=250 \). Since \( 2^7\lt250 \lt 2^8 \) and \( 250 \) is closer in value to \( 256 \) than \( 128 \), we check,\[\begin{align*} 2^{7.9} &\approx 238.8564 \\ 2^{7.95} &\approx 247.2797 \\ 2^{7.97} &\approx 250.7316 \\ 2^{7.96} &\approx 248.9997 \end{align*}\] Therefore, \( \log_2 (250) \approx 7.97 \) to two decimal places of accuracy.

1. Using the following tables of values,

   \( x \)	\( A(x) = \log_{\tfrac{1}{5}} {\left( x \right)} \)
   \( \tfrac{1}{5} \)	\( 1 \)
   \( 1 \)	\( 0 \)
   \( 5 \)	\( -1 \)
   \( 25 \)	\( -2 \)

   \( x \)	\( B(x) = \log_{\tfrac{1}{2}} {\left( x \right)} \)
   \( \tfrac{1}{2} \)	\( 1 \)
   \( 1 \)	\( 0 \)
   \( 2 \)	\( -1 \)
   \( 4 \)	\( -2 \)

   \( x \)	\( C(x) = \log_{3} {\left( x \right)} \)
   \( \tfrac{1}{3} \)	\( -1 \)
   \( 1 \)	\( 0 \)
   \( 3 \)	\( 1 \)
   \( 9 \)	\( 2 \)

   \( x \)	\( D(x) = \log_{8} {\left( x \right)} \)
   \( \tfrac{1}{8} \)	\( -1 \)
   \( 1 \)	\( 0 \)
   \( 8 \)	\( 1 \)
   \( 64 \)	\( 2 \)

   produces the sketch

2. All four functions have a domain \( \{x\in\mathbb{R} \mid x \gt 0\} \), a range \( \{y\in\mathbb{R}\} \), an \( x \)-intercept at \( (1,0) \), and a vertical asymptote of \( x=0 \). Since \( A(x) \) and \( B(x) \) are logarithmic functions with a base less than \( 1 \), they are decreasing functions. Since \( C(x) \) and \( D(x) \) have a base greater than \( 1 \), they are increasing functions.
3. Since \( \frac{1}{5}\lt\frac{1}{2}\lt\frac{2}{3} \lt 1 \), the function \( f(x)=\log_\frac{2}{3} (x) \) is an decreasing function that will lie above \( B(x)=\log_\frac{1}{2} (x) \) when \( 0\lt x \lt1 \) and below \( B(x)=\log_\frac{1}{2} (x) \) when \( x \gt 1 \), intersecting all four functions at \( (1,0) \). Since \( 1 \lt 3\lt 5 \lt 8 \), the function \( g(x)=\log_5 (x) \) is a increasing function that will lie between the graphs of the functions \( C(x)=\log_3 (x) \) and \( D(x)=\log_8 (x) \), intersecting all four functions at \( (1,0) \).

1. To find the inverse of \( g(x)=3^{2(x-1)}+5 \), interchange \( x \) and \( y \) to obtain \( x=3^{2(y-1)}+5 \).

   Now, solving for \( y \),

   \[\begin{align*} x-5 &= 3^{2(y-1)} \\ 2(y-1) &= \log_3(x-5) \\ y-1 &= \frac{1}{2} \log_3(x-5) \end{align*}\]

   Therefore, \( g^{-1}(x) = \frac{1}{2} \log_3(x-5)+1 \).

2. To find the inverse of \( g(x)=2 \log_5(x+4)-1 \), interchange \( x \) and \( y \) to obtain \( x=2\log_5(y+4)-1 \).

   Now, solving for \( y \),

   \[\begin{align*} x+1 &= 2\log_5(y+4) \\ \frac{1}{2}(x+1) &= \log_5(y+4) \\ y+4 &= 5^{\frac{1}{2}\left(x+1\right)} \end{align*}\]

   Therefore, \( g^{-1}(x)=5^{\frac{1}{2}\left(x+1\right)}-4 \).

3. To find the inverse of \( g(x)=-\log_3(\frac{1}{2}x-3)+4 \), interchange \( x \) and \( y \) to obtain \( x=-\log_3(\frac{1}{2}y-3)+4 \).

   Now, solving for \( y \),

   \[\begin{align*} x-4 &= -\log_3(\frac{1}{2}y-3) \\ 4-x &= \log_3(\frac{1}{2}y-3) \\ \frac{1}{2}y-3 &= 3^{4-x} \\ \frac{1}{2}y &= 3^{4-x}+3 \end{align*}\]

   Therefore, \( g^{-1}(x)=2\left(3^{4-x}\right)+6 \).

The graph of \( f(x)=2\log_2(x-2) \) can be obtained by applying the following transformations to the graph of \( f(x)= \log_2(x) \)

• a vertical stretch about the \( x \)-axis by a factor of \( 2 \)
• a horizontal translation right \( 2 \) units

The graph of \( g(x)=2-\log_3(x-3)=-\log_3(x-3)+2 \) can be obtained by applying the following transformations to the graph of \( f(x)= \log_3(x) \)

• a reflection in the \( x \)-axis
• a horizontal translation right \( 3 \) units
• a vertical translation up \( 2 \) units

The two functions appear to intersect at \( x=4 \). To verify this, show that \( f(4)=g(4) \).

Therefore, \( f(x)=2\log_2(x-2) \) and \( g(x)=2-\log_3(x-3) \) intersect at \( \left(4, 2\right) \).