anonymous
  • anonymous
Log4 X-log16 (x+3)=1/2 The 4 and 16 are bases can someone please tell me how to solve this?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
change of base perhaps??
amistre64
  • amistre64
4^y = x 16^y-3 = x would that work out ?
anonymous
  • anonymous
change both bases to 4, then use the laws of exponents to combine\[\log_{16}(x+3)=y\] means \[16^y=(x+3)\] \[4^{2y}=x+3\] \[2y=log_4(x+3)\] \[y=\frac{log_4(x+3)}{2}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
4^y - 4^(2y-6) = 1/2
amistre64
  • amistre64
i was close lol
anonymous
  • anonymous
change of base gets me two joined equations with a factored out x
amistre64
  • amistre64
log4(x) - [log4(x+3)/log4(16)] = 1/2 log4(x) - (1/2)log4(x+3) = 1/2
anonymous
  • anonymous
idk if i can change both bases to 4
amistre64
  • amistre64
log4(x/sqrt(x+3)) = 1/2
anonymous
  • anonymous
you can do that?
amistre64
  • amistre64
sqrt(4) = sqrt(x/(x+3)) i believe so :)
anonymous
  • anonymous
then exponentiate to get \[\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}\]
amistre64
  • amistre64
4 = x/x+3 4x+12 = x 3x = -12 x = -4...maybe
amistre64
  • amistre64
i got lost in it for a moment....
anonymous
  • anonymous
no way -4 is an answer since you cannot take log of -4
amistre64
  • amistre64
i know i know.... my brain took a momentary leave of absence lol
anonymous
  • anonymous
the 4 is not negative
amistre64
  • amistre64
sqrt(4) = x/(x+3) is what i meant to type :)
amistre64
  • amistre64
4 = x^2/x^2+9+6x
amistre64
  • amistre64
4x^2 +24x +36 = x^2 3x^2 +24x +36 = 0... thats better me thinks
amistre64
  • amistre64
3(x^2 +8x +12) = 0
amistre64
  • amistre64
x = -6 and -2 by that accounting...... so maybe not ;)
anonymous
  • anonymous
oops start with \[\frac{x}{\sqrt{x+3}}=4^{\frac{1}{2}}\]
amistre64
  • amistre64
...... yeah, it would help if I did the actual problem instead of making up me own lol
anonymous
  • anonymous
\[x=2\sqrt{x+3}\]
anonymous
  • 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.
anonymous
  • anonymous
yup 6 it is
anonymous
  • anonymous
hey where did the log16 go?
amistre64
  • amistre64
we..and be that i me sat.... modified it to a log4(x+3)/2
anonymous
  • anonymous
I get it its just i do not know where the log16 went the log of16 is 1.20411
amistre64
  • amistre64
change of base just changes the appearance and not the value
amistre64
  • amistre64
log4(16) not log(16)
anonymous
  • anonymous
let us go slow. \[log_{16}(x)=y\] means \[16^y=x\] for example \[log_{16}(4096)=3\] because \[16^3=4096\]
anonymous
  • anonymous
but \[16=4^2\] so it is also true that \[(4^2)^3=4069\] \[4^6=4069\] so \[log_4(4069)=6\]
anonymous
  • anonymous
so the log base 16 did not disappear, just converted to log base 4
anonymous
  • anonymous
oho i get it thank you :]
anonymous
  • anonymous
i got X=2.6667
anonymous
  • anonymous
how did you get that? i am fairly certain x = 6

Looking for something else?

Not the answer you are looking for? Search for more explanations.