## King 3 years ago Its a log question so i shall give it in the reply......

1. King

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

2. FoolForMath

alright what have you tried ?

3. King

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

4. Stom

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

5. Lumenaire

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

6. King

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

7. Lumenaire

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

8. FoolForMath

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

9. King

no log x base y

10. King

xplain

11. Lumenaire

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

12. Lumenaire

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

13. King

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

14. King

oh ok

15. King

so ok how to find x,y

16. Lumenaire

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

17. FoolForMath

yes king

18. FoolForMath

king its just substitution in the rest parts ..

19. Lumenaire

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

20. King

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

21. Lumenaire