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 .