Question

another log question

Answer 1

\[\log_{y}x+\log_{x}y=\frac{ 10 }{ 3 } \]

xy=144

compute (x+y)/2

Answer 2

what have you tried?

Answer 3

if you take \(\log_xy =a\)
you'll get a quadratic in a,

Answer 4

oh

Answer 5

logy(x) is simply the inverse right?

Answer 6

log_x y = 1/ log_y x

Answer 7

"reciprocal", not inverse :)

Answer 8

so now that i have a+(1/a)=10/3

i can multiply by a?

Answer 9

yep.

solve that quadratic.

Answer 10

after i solve the quadratic, i would that the answers back to log(y)x?

Answer 11

so, you'll have
\(\log_x y = \dfrac{\log y}{\log x}\)

and since xy = 144
\(\log xy = 144 \\ \log x + \log y = 144\)

we have log y/log x, log x +log y
2 equations, 2 unknowns!

Answer 12

****
\(\log x +\log y = \log 144\)

Answer 13

the quadratic i got 3 and 1/3

Answer 14

so your 2 equations will solve for x and y?

Answer 15

for log x and log y
and hence yes, for x and y

Answer 16

log y = 3 log x
log x + log y = log 144


log x + 3log x = 144
4 log x = log 144

log x = ...
x = ...

Answer 17

where did the 3logx come from

Answer 18

"log y = 3 log x"

Answer 19

how did you come to that statement?

Answer 20

\(\log_xy =3 \\ \dfrac{\log y}{\log x} = 3 \\ \log y = 3 \log x\)

Answer 21

4logx=log144

is that equal to logx=log36?

i probably messed up a rule here

Answer 22

yeah..

log x = (log 144)/4

\(\Large \log 144^{\dfrac{1}{4}}\)

Answer 23

so x is -0.279?

Answer 24

negative?

Answer 25

0.53 i forgot 1/4 isnt really the exponent

Answer 26

we can keep it as \(\sqrt {12}\)

Answer 27

now find log y and hence y

log y = 3 log x = ...

Answer 28

3(sqrt12)

Answer 29

log x is not sqrt 12
x is sqrt 12

Answer 30

\(\log y =3 \log x = 3 \log (144^{0.25} ) = 3 \log \sqrt {12} = \log (\sqrt{12}^3) \\ \implies y = \sqrt{12}^3 \)

Answer 31

see if you get that?

Answer 32

oh yeah

Answer 33

so 12(sqrt12)

Answer 34

yes

Answer 35

final answer is 13(sqrt3)

Answer 36

yes

Answer 37

alright thanks

Answer 38

welcome ^_^