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


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

$Yes sir, but it is the workings i {don}^{'} t know sir .$
Commented by mr W last updated on 19/Jul/20

$i ‘ ‘ {s e e}^{″} t h e s o l u t i o n r = 2, p = 9 .$ $i t i s e x p e c t e d t h a t y o u c a n s e e t h e$ $s o l u t i o n, n o t c a l c u l a t e . i f t h e s e c o n d$ $e q n . i s r + p = 12, y o u c a n n o t s e e a n d$ $a l s o n o t c a l c u l a t e t h e s o l u t i o n .$
Commented by I want to learn more last updated on 19/Jul/20

$Alright . Thanks sir .$
	Answered by 1549442205PVT last updated on 19/Jul/20
	$Put log_r 8=a,log_3 p=b.We get 8=r^a ,p=3^b (r>0,r≠1,p>0) { ((a+b=5(1))),((3^b +8^(1/a) =11(2))) :} From (1) we get b=5−a ,replace into (2) we get 3^(5−a) +8^(1/a) −11=0(1).we need to have r=8^(1/a) <11⇒a>((ln8)/(ln11))≈0.8671. Putting f(a)=3^(5−a) +8^(1/a) −11.It is easy to see that f(3)=0,so a=3 is root of eqs.(1) we will prove it is unique root.Indeed, f ′(a)=−ln3×3^(5−a) +8^(1/a) (−(1/a^2 ))<0∀a∈(((ln8)/(ln11));+∞) hence a=3 is unique root of eqs.(1) ⇒b=2.From that we get r=2,p=9$
	$Put \log_{r} 8 = a, \log_{3} p = b . We get$ $8 = r^{a}, p = 3^{b} (r > 0, r \neq 1, p > 0)$ ${\begin{cases} a + b = 5 (1) \\ 3^{b} + 8^{\frac{1}{a}} = 11 (2) \end{cases}$ $From (1) we get b = 5 - a, replace into$ $(2) we get 3^{5 - a} + 8^{\frac{1}{a}} - 11 = 0 (1) . we need to$ $have r = 8^{\frac{1}{a}} < 11 \Rightarrow a > \frac{\ln 8}{\ln 11} \approx 0.8671 .$ $Putting f (a) = 3^{5 - a} + 8^{\frac{1}{a}} - 11 . It is easy to see$ $that f (3) = 0, so a = 3 is root of eqs . (1)$ $we will prove it is unique root . Indeed,$ $f^{'} (a) = - \ln 3 \times 3^{5 - a} + 8^{\frac{1}{a}} (- \frac{1}{a^{2}}) < 0 \forall a \in (\frac{\ln 8}{\ln 11}; + \infty)$ $hence a = 3 is unique root of eqs . (1)$ $\Rightarrow b = 2 . From that we get r = 2, p = 9$
	Commented by I want to learn more last updated on 19/Jul/20

	$I appreciate sir .$
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