The functions are
and
.
(a)
\Solve
.
Substitute
.

Apply logarithm one to one property : if
,then
.


.
Therfore,the solution is
.
The solution set is
.
The point is on the graph of
is
.
(b)
\Solve
.
Substitute
.

Apply logarithm one to one property : if
,then
.




Therefore,the solution is
.
The solution set is
.
The point is on the graph of
is
.
(c)
\Solve for
.
Equate
and
.

Apply logarithm one to one property : if
,then
.




Therefore, the solution is
.
The solution set is 
Substitute
in
.




Therefore,the point is
.
Yes, the graphs of the functions
and
intersect at the point
.
(d)
\Solve for
.
.
Substitute
and
.

Apply logarithm product property :
.

Apply logarithm one to one property : if
,then
.




Solve for
by factorizing the quadratic equation.



Apply zero product rule.
\
or 
or 
Since
does not exist, hence it is not considered.
Therefore, the solution is
.
The solution set is
.
(e)
\Solve for
.

Substitute
and
.

Apply logarithm quotient property :
.

Apply logarithm one to one property : if
,then
.

Solve for
.




Therefore,the solution is
.
The solution set is
.
(a) The solution set is
and the point is on the graph of
is
.
(b) The solution set is
and the point is on the graph of
is
.
(c) The solution set is
.
Yes,the grpahs of the functions
and
are intersected at the point
.
(d) The solution set is
.
(e) The solution set is
.