Question

Its a log question so i shall give it in the reply......

Answer 1

\[\log_{3} (\log_{2} x) + \log_{1/3} (\log_{2} y)=1.\]
Find x,y if given \[xy ^{2}=4\]

Answer 2

alright what have you tried ?

Answer 3

um i took log base 1/3 as log base 3^-1 and that will be
\[-\log_{3} \]

Answer 4

log3(log2,x) + log1/3(log2,y) =1

log3(log2,x) - log3(log2,y) =1

log3(log2,x/log2,y) =1
log3(x/y) =1
x/y=3

Answer 5

replace 1 with log base 3 of 3
then:
\[\log_{3}(\log_{2}x) - \log_{3}(\log_{2}y) = \log_{3}3 \]
then:\[\log_{2}x / \log_{2}y = 3\]
\[\log_{y}x = 3 \]
y^3 = x
so y is 4^0.2
x = 4^0.6

Answer 6

can u please xplain how u got the 3rd equation?

Answer 7

log a - log b = log(a/b)

Answer 8

@King:From the first equation I got $\frac{x}{y} = 3$ can you do the rest ?

Answer 9

no log x base y

Answer 10

xplain

Answer 11

oh ok. log a / log b = log a base b

Answer 12

if they both have the same base. that is the rule

Answer 13

FFM what have u written?$\frac{x}{y} = 3$

Answer 14

oh ok

Answer 15

so ok how to find x,y

Answer 16

read the rest. y^3 is x. and just plug that into your other equation

Answer 17

yes king

Answer 18

king its just substitution in the rest parts ..

Answer 19

xy^2 becomes y^5
so y^5 =4 so y = 4^0.2
and x = y^3 = 4^0.6

Answer 20

ok...thnx guys can u solve 1 more question....shall i post it?

Answer 21

go ahead