Question Number 104086 by I want to learn more last updated on 19/Jul/20

Solve:      log_r 8   +   log_3 p   =  5         ..... (i)                             r   +  p   =  11         ..... (ii)

Commented byI want to learn more last updated on 19/Jul/20

Yes sir, but it is the workings i don′t know sir.

Commented bymr W last updated on 19/Jul/20

i ♮seeε the solution r=2, p=9.  it is expected that you can see the  solution, not calculate. if the second  eqn. is r+p=12, you can not see and  also not calculate the solution.

Commented byI want to learn more last updated on 19/Jul/20

Alright. Thanks  sir.

Answered by 1549442205PVT last updated on 19/Jul/20

Put log_r 8=a,log_3 p=b.We get  8=r^a ,p=3^b  (r>0,r≠1,p>0)   { ((a+b=5(1))),((3^b +8^(1/a) =11(2))) :}  From (1) we get b=5−a ,replace into  (2) we get 3^(5−a) +8^(1/a) −11=0(1).we need to  have r=8^(1/a) <11⇒a>((ln8)/(ln11))≈0.8671.  Putting f(a)=3^(5−a) +8^(1/a) −11.It is easy to see  that f(3)=0,so a=3 is root of eqs.(1)  we will prove it is unique root.Indeed,  f ′(a)=−ln3×3^(5−a) +8^(1/a) (−(1/a^2 ))<0∀a∈(((ln8)/(ln11));+∞)  hence a=3 is unique root of eqs.(1)  ⇒b=2.From that we get r=2,p=9

Commented byI want to learn more last updated on 19/Jul/20

I appreciate sir.