Question

If x,y > 0, log_y (x) + log_x (y) = 10/3 and xy = 144, then (x + y)/2 =

Answer choices: available upon request.
(How do you get the answer?)

Answer 1

i will byte u :P

Answer 2

\[\left( \log_yx\right)^2 + 1 = \frac{10}3\log_yx\]Complete squares to get the value of \(\log_yx\).

Answer 3

ops bite :p

Answer 4

@Ishaan94 is right :)
Solve using quadratic formula the equation he wrote

Answer 5

log_y (x) = log_x (x)/log_x(y)
so:
log_x (x)/log_x(y) +log_x(y) = 10/3
log_x(x/y *y) =10/3
x=10/3
rest should be easy

Answer 6

@myko , log_x (x)=1

Answer 7

:) good point

Answer 8

@myko , if you substitute that, you will get same equation as @Ishaan94 got

Answer 9

ishaan, how did you get that?

Answer 10

but i think my way is more direct

Answer 11

@ByteMe , follow @myko 's method to get @Ishaan94 's equation

Answer 12

@myko you have made a mistake

Answer 13

log_a(b)=a/b

Answer 14

log_x(y)+log_y(x)=y/x+x/y=

Answer 15

@myko
\[{{\log_x (y)} \over {\log_x (z)} }\neq \log_x{{y} \over {z}}\]

Answer 16

log_x(y)=only y/x

Answer 17

now simplify :P

Answer 18

first of all you need to understand the change of base formula
\[\large \log_{x}y = \frac{\log y}{\log x} = \frac{1}{\frac{\log x}{\log y}} = \frac{1}{\log_{y}x} \]
Let a = log_y (x)
\[\rightarrow a + \frac{1}{a} = \frac{10}{3}\]
\[\rightarrow a^{2} +1 = \frac{10}{3}a\]
(ishaans equation)

Answer 19

y/x+x/y=10/3
=>y^2+x^2/xy=10/3

Answer 20

yo whatever :P

Answer 21

@dumbcow , very good explanation .
@myko , Now you are right :)

Answer 22

ya, you right...:)
log_y (x) = log_x (x)/log_x(y)
so:
log_x (x)/log_x(y) +log_x(y) = 10/3
1/log_x(y) +log_x(y)=10/3
log_x(y)^2 +1=10/3log_x(y)^2

Answer 23

So i'm gettng a=14/3 and a=2 from completing the square. How do i get (x+y)/2 ?

Answer 24

@ByteMe , check again. I am getting

\[a=3\]
and
\[a={1 \over 3}\]

Answer 25

i agree, this implies there are 2 answers for (x+y)/2

Answer 26

ok ill check. theres only one answer though. Do I put it up?

Answer 27

Case 1. a= 3
a=logx(y)=3

y=x^(3)

Now use x*y = 144

144/x=x^3

144=x^4

x= ±√12

Now substitute this in x*y=144 to get your y

Case 2, a= 1/3
Proceed above as in case 1

Answer 28

nvmd. i see where i made my mistake.

Answer 29

i stand corrected...either case yields equivalent x,y values

Answer 30

@dumbcow ,yes

so final answer I am getting
\[{13\sqrt{12} }\over 2\]

Answer 31

On simplifying it becomes
or \[13\sqrt(3)\]

Answer 32

yes!!! thank you very much guys!!!

Answer 33

@dumbcow , I am not certain about the sign of x and y.

Answer 34

i'm sorry i have only 1 medal to give :(

Answer 35

inee meenie mynee moe!!!

Answer 36

both x,y are positive

Answer 37

Yes ofcourse :P It is log :)

Answer 38

I meant base of log and the quantity inside log has to be positive :)

Answer 39

haha :)

Answer 40

you guys are the best!

Answer 41

Nope we are just learning while sharing :)

Answer 42

Nope we are just learning while sharing :)

Answer 43

Nope we are just learning while sharing :)