Question 969212
Interesting problems.  Here is a way to deal with #3.


Let {{{v=x^(1/2)}}}.  This means {{{x^(1/4)=v^2}}}.


{{{v+3v^2=10}}}
{{{3v^2+v-10=0}}}
discriminant is {{{1^2+4*3*10=1+120=121=11^2}}}.


{{{v=(-1+- sqrt(11^2))/(2*3)}}}
{{{v=(-1+- 11)/6}}}
{{{v=-2}}} or {{{v=5/3}}}


Substitute back.
{{{x^(1/2)=-2}}}  or  {{{x^(1/2)=5/3}}}

{{{highlight(x=4)}}}  OR  {{{highlight(x=25/9)}}}.




Just a start in attempting #2:


{{{2*log(3,(81)^(1/3))+e^(-4ln(t))}}}


{{{2*log(3,((3^4)^(1/3)))+e^(-4ln(t))}}}


{{{2*log(3,(3^(4/3)))+e^(-4ln(t))}}}


{{{2(4/3)*log(3,3)+e^(-4ln(t))}}}


{{{8/3+e^(-4*ln(t))}}}


...think what you can do with that last term with e....
...work with the exponent.


{{{8/3+e^(ln(t^(-4)))}}}, you need to recognize the rule of logarithms used in the step. 
Now recognize ln() function and e^() function are inverses of each other.


{{{8/3+t^(-4)}}}


{{{highlight(8/3+1/t^4)}}}