Question

xyz = 2^6 and (2 log x)( 2 log yz) + (2log y)(2 log z) = 10 with x,y,z >= 0. Then sqrt(2 log^2 x + 2 log^2 y + 2 log^2 z) = ....

Answer 1

\[xyz = 10^6\]
and
\[(\log _{2} x) (\log _{2} yz) + (\log _{2} y) (\log _{2} z) = 10 \]
with x,y,z >= 0

Then
\[\sqrt{\log_{2}^{2}x + \log _{2}^{2}y + \log_{2}^{2}z} = ....\]

Answer 2

@agent0smith @3mar

Answer 3

Where you got such problems, Kevin?

Answer 4

all. lol

Answer 5

you need x and y and z. right?

Answer 6

The answer is 4, maybe you can explain to me about how to get that

Answer 7

no... find sqrt of bla bla bla as I wrote above

Answer 8

I see

Answer 9

I will do my best In sha' Allah

Answer 10

Let's begin.

Answer 11

thx :D

Answer 12

You are welcome!

Answer 13

Do you have Skype Id?

Answer 14

It will be more easier there!

Answer 15

@Directrix @IrishBoy123 @Astrophysics @SapphireMoon @sweetburger @jackthegreatest

Answer 16

Sorry, but I don't have :(

Answer 17

Hi @Will.H

Answer 18

for
\[xyz=10^6\]
take the log to the base 2 for both sides
\[\log_{2} (xyz)=\log_{2}(10^6) \]

Answer 19

then multiply both sided to 2
\[2\log_2(xyz)=2\log_210^6\]

Answer 20

As you know the feature of the exponent of the logarithmic function:
\[alogx=\log(x^2)\]

Answer 21

sorry
\[a*\log(x)=\log(x^a)\]

Answer 22

Do you follow?

Answer 23

yup

Answer 24

just go on :)

Answer 25

then it would take the form:
\[\log_2(xyz)^2=2\log_210^6\]

then distribute the power 2 to x, y and z
\[\log_2(x^2.y^2.z^2)=2\log_210^6\]

Answer 26

then separete what inside the log() by plus sign as follows:
\[\log_2(x^2)+\log_2(y^2)+\log_2(z^2)=2\log_210^6\]

then take the square root for both sides:
\[\sqrt{\log_2(x^2)+\log_2(y^2)+\log_2(z^2)}=\sqrt{2\log_210^6}=number\]

That's it.

Done!

Answer 27

Hope that helps

Answer 28

got it or stick?

Answer 29

actually the question is sqrt of 2 log 2 x + blablabla....
not sqrt of 2 log (x^2) .....
Is it just me that I don't understand?
Thx anyway, I've figured out how is it after following your first method on the post above. I got stick at that before.

Answer 30

It's like this...