## anonymous 5 years ago Log4 X-log16 (x+3)=1/2 The 4 and 16 are bases can someone please tell me how to solve this?

1. amistre64

change of base perhaps??

2. amistre64

4^y = x 16^y-3 = x would that work out ?

3. anonymous

change both bases to 4, then use the laws of exponents to combine$\log_{16}(x+3)=y$ means $16^y=(x+3)$ $4^{2y}=x+3$ $2y=log_4(x+3)$ $y=\frac{log_4(x+3)}{2}$

4. amistre64

4^y - 4^(2y-6) = 1/2

5. amistre64

i was close lol

6. anonymous

change of base gets me two joined equations with a factored out x

7. amistre64

log4(x) - [log4(x+3)/log4(16)] = 1/2 log4(x) - (1/2)log4(x+3) = 1/2

8. anonymous

idk if i can change both bases to 4

9. amistre64

log4(x/sqrt(x+3)) = 1/2

10. anonymous

you can do that?

11. amistre64

sqrt(4) = sqrt(x/(x+3)) i believe so :)

12. anonymous

then exponentiate to get $\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}$

13. amistre64

4 = x/x+3 4x+12 = x 3x = -12 x = -4...maybe

14. amistre64

i got lost in it for a moment....

15. anonymous

no way -4 is an answer since you cannot take log of -4

16. amistre64

i know i know.... my brain took a momentary leave of absence lol

17. anonymous

the 4 is not negative

18. amistre64

sqrt(4) = x/(x+3) is what i meant to type :)

19. amistre64

4 = x^2/x^2+9+6x

20. amistre64

4x^2 +24x +36 = x^2 3x^2 +24x +36 = 0... thats better me thinks

21. amistre64

3(x^2 +8x +12) = 0

22. amistre64

x = -6 and -2 by that accounting...... so maybe not ;)

23. anonymous

oops start with $\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}$

24. amistre64

...... yeah, it would help if I did the actual problem instead of making up me own lol

25. anonymous

$x=2\sqrt{x+3}$

26. anonymous

$x^2=4(x+3)=4x+12$ $x^2-4x-12=0$ $(x-6)(x+2)=0$ $x=6, x=-2$ -2 not a solution. we can try 6 to see if it works.

27. anonymous

yup 6 it is

28. anonymous

hey where did the log16 go?

29. amistre64

we..and be that i me sat.... modified it to a log4(x+3)/2

30. anonymous

I get it its just i do not know where the log16 went the log of16 is 1.20411

31. amistre64

change of base just changes the appearance and not the value

32. amistre64

log4(16) not log(16)

33. anonymous

let us go slow. $log_{16}(x)=y$ means $16^y=x$ for example $log_{16}(4096)=3$ because $16^3=4096$

34. anonymous

but $16=4^2$ so it is also true that $(4^2)^3=4069$ $4^6=4069$ so $log_4(4069)=6$

35. anonymous

so the log base 16 did not disappear, just converted to log base 4

36. anonymous

oho i get it thank you :]

37. anonymous

i got X=2.6667

38. anonymous

how did you get that? i am fairly certain x = 6