anonymous
  • anonymous
Solve. {log(1/2)[4x]}^2+log(1/2)[x^2/8]=8
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1332961406364:dw|
anonymous
  • anonymous
\[\log_{\frac{1}{2}}4x+\log_{\frac{1}{2}}\frac{x^2}{8}=8?\]
anonymous
  • anonymous
The first one is squared and second base is 2 not 1/2.

Looking for something else?

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

More answers

anonymous
  • anonymous
is 4x squared or the logbase(1/2) 4x is squared?
anonymous
  • anonymous
logbase(1/2) 4x is squared.
anonymous
  • anonymous
so we'll do a change of base and solve a quadratic
anonymous
  • anonymous
\[\log_{\frac{1}{2}}4x=\frac{\log_{2}4x}{\log_{2}\frac{1}{2}}\]
anonymous
  • anonymous
\[\log_{2}\frac{1}{2}=-1\]
anonymous
  • anonymous
so we have \[(-\log_{2} 4x)^2+\log_{2}\frac{x^2}{8}=8\]
anonymous
  • anonymous
now we will represent 4 and 1/8 and powers of 2
anonymous
  • anonymous
\[(-\log_{2}2^2x)^2+\log_{2}2^{-3}x=8\]
anonymous
  • anonymous
use properties of logs to expand the logs using addition
anonymous
  • anonymous
\[[-(2\log_{2}2+\log_{2}x)]^2+[-3\log_{2}2+\log_{2}x]=8\]
anonymous
  • anonymous
\[-(2+\log_{2}x)^2-3+\log_{2}x=8\]
anonymous
  • anonymous
Just wanted to ask if there was a mistake, but you managed to fix it already. :)
anonymous
  • anonymous
yes it is a big pain in the butt using this equation editor because the copy and paste is broken
anonymous
  • anonymous
Understood.
anonymous
  • anonymous
\[-(4+4\log_{2}x+(\log_{2}x)^2)-3+\log_{2}x=8\]now we expand into ax^2+bx+c form
anonymous
  • anonymous
\[-4-4\log_{2}x-(\log_{2}x)^2-3+\log_{2}x=8\]
anonymous
  • anonymous
\[-(\log_{2}x)^2-4\log_{2}x-7=8\]
anonymous
  • anonymous
let\[u=\log_{2}x\] \[-u^2-4u-7=8\] \[u^2+4u+7=-8\] \[u^2+4u+15=0\]
anonymous
  • anonymous
hmm, maybe i made a mistake somewhere
anonymous
  • anonymous
ahh i see i forgot to square the x in the 2nd logarithm
anonymous
  • anonymous
its x^2/8
anonymous
  • anonymous
\[[-(2\log_{2}2+\log_{2}x)]^2+[-3\log_{2}2+2\log_{2}x]=8\]
anonymous
  • anonymous
look right now?
anonymous
  • anonymous
At the moment, yes.
anonymous
  • anonymous
\[[-(2+\log_{2}x)]^2-3+2\log_{2}x=8\] lets see if we get somewhere with this
anonymous
  • anonymous
\[-(4+4\log_{2}x+(\log_{2}x)^2)-3+2\log_{2}x=8\] \[-(\log_{2}x)^2-4\log_{2}x-4-3+2\log_{2}x=8\] \[-u^2-2u-15=0\] dammit, another mistake? lol
anonymous
  • anonymous
should've used paper
anonymous
  • anonymous
I know the feeling, just fail to do it.
anonymous
  • anonymous
let me grab some paper, check back in a few mins
anonymous
  • anonymous
ahh it was all a stupid negative sign
anonymous
  • anonymous
What's the quadratic supposed to be if you replace \[\log_{2}x\] with u?
anonymous
  • anonymous
i've got u^2+2u-7=0
anonymous
  • anonymous
can u draw the problem out again so i can be sure i'm doing the right problem?
anonymous
  • anonymous
|dw:1332964583627:dw|
anonymous
  • anonymous
ok i finally got an answer that makes sense
anonymous
  • anonymous
u^2+6u-7=0
anonymous
  • anonymous
Thank you so much!
anonymous
  • anonymous
you want me to show steps?
anonymous
  • anonymous
I can do it on my own from now on. :)
anonymous
  • anonymous
you still have to resubstitute log_2(x) for u
anonymous
  • anonymous
ok cool, sorry for themistakes, should've used paper from the start
anonymous
  • anonymous
Don't worry about the mistakes, eventually everything turned out the way needed. Thank you again!
anonymous
  • anonymous
yw

Looking for something else?

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