Question

Urgent!!!!!!
solve for x :
log^{2}_{1/2} 4x + log_{2} x^2/8 = 8

Answer 1

\[\log^{2}_{1/2} 4x + \log_{2} x^2/8 = 8 \]

Answer 2

does log^2 even exist?

Answer 3

i don't know, its in my HW

Answer 4

check the statement again?

Answer 5

its correct

Answer 6

and can you please re-post the ques,,using brackets this time..

Answer 7

\[\log^{2}_{1/2} 4x + \log_{2} x^2/8 = 8 \]

Answer 8

wait i ll try to solve

Answer 9

break log(4x) = log 4 + log x
since base is 1/2 here,,we have logx -1/2= -1/log x -1/2 where base has been converted to 2
expanding other term,,we have 2 logx -3
ultimately,,on simplifying,,we will have a cubic eqn..hmmn..maybe any other mthod..?

Answer 10

hey so @shubham.bagrecha is the way you have your problem exactly the way it appears in your book?

Answer 11

yes

Answer 12

In my book there are 3 such questions.

Answer 13

pls anyone help!!!!!!!!

Answer 14

\[(\log_\frac{1}{2}(4x))^2+\log_2(\frac{x^2}{8})=8\]

So we know the rules log(ab)=log(a)+log(b) and log(a/b)=log(a)-log(b)
So using those we get
\[(\log_\frac{1}{2}(4)+\log_\frac{1}{2}(x))^2+\log_2(x^2)-\log_2(8)=8\]

I'm gonna use this rule log(a^r)=r log(a)

\[(\log_\frac{1}{2}(4))^2+2 \cdot \log_\frac{1}{2}(4) \log_\frac{1}{2}(x)+(\log_\frac{1}{2}(x))^2+2 \log_2(x)-\log_2(8)=8\]
So hmm... this looked like it didn't help

Answer 15

What class are you in school?

Answer 16

11

Answer 17

and you?

Answer 18

I think you are looking at your problem wrong.
This is like a beginning algebra class right?
What book are you looking at?

Answer 19

this book don't have any publisher or any author.

Answer 20

CBSE? NCERT ?

Answer 21

no

Answer 22

hmm..buddy it'll form a cubic eqn,,which altough,,is solvable but i dont guess its what you'll be looking for..

Answer 23

I'm going to take a wild guess
This is a 11 grade problem
Do you mean for log parts to have the same base
Is this the problem:
\[\log_2(4x)^\frac{1}{4}+\log_2(\frac{x^2}{8})=8\]
?

Answer 24

\[\log_2(4x)^\frac{1}{2}+\log_2(\frac{x^2}{8})=8 ?\]

Answer 25

no i've asked the right question.

Answer 26

it has a log-square

Answer 27

ok thanks all for whatever support you've given.

Answer 28

Ok what I think I might have something

Answer 29

\[\log_\frac{1}{2}(u)=\frac{\ln(u)}{\ln(\frac{1}{2})} \text{ by change of base}\]
\[=\frac{\ln(u)}{\ln(1)-\ln(2)}=\frac{\ln(u)}{0-\ln(2)}=\frac{\ln(u)}{-\ln(2)}=-\frac{\ln(u)}{\ln(2)}\]
\[=-\log_2(u) \text{ by change of base again }\]

So we have
\[(-\log_2(4x))^2+\log_2(\frac{x^2}{8})=8\]

But we still have that log that is squared

Answer 30

\[(\log_2(4x))^2+\log_2(\frac{x^2}{8})=8\]

Answer 31

but now i'm stuck here
still thinking

Answer 32

ohh damn..i used the same thing @myininaya ,,but i did wrong in 1 step..
you're right,,it'll be a quad eqn now now expanding..hmm

Answer 33

Oh!

Answer 34

Going to use log(ab)=log(a)+log(b)
\[(\log_2(4)+\log_2(x))^2+\log_2(\frac{x^2}{8})=8\]
using log(a/b)=log(a)-log(b)
\[(\log_2(4))^2+2 \cdot \log_2(4) \log_2(x)+(\log_2(x))^2+\log_2(x^2)-\log_2(8)=8\]
I did the above using (a+b)^2=a^2+2ab+b^2

now using log(a^r)=r log(a)
We have

\[(\log_2(4))^2+2 \cdot \log_2(4) \log_2(x)+(\log_2(x))^2+2 \log_2(x)-\log_2(8)=8\]

Now gathering like terms and subtracting 8 on both sides we have

\[(\log_2(x))^2+2\log_2(x)+2 \log_2(4) \log_2(x)-\log_2(8)+(\log_2(4))^2-8=0\]

Answer 35

\[\log_2(4)=2\]
so we have...

Answer 36

and \[\log_2(8)=3 \]

so we have...
\[(\log_2(x))^2+2\log_2(x)+2 (2) \log_2(x)-3+(2)^2-8=0\]

Answer 37

So combine those constants
and multiply a little

Answer 38

alright @myininaya you did an amazing job,,now let him try a bit also :)

Answer 39

\[(\log_2(x))^2+2\log_2(x)+4 \log_2(x)-7=0 \]
Of all his now ... :)

Answer 40

ok*