Question

help pls:)

Answer 1

Generally we need a question to help with

Answer 2

\[\log_{10} x-\log_{10} \sqrt{x} = 2/(\log_{10}x)\]

Answer 3

this is the question sorry for late reply, danglobery now u should help me , there is question now

Answer 4

help me any1

Answer 5

ok let me try

Answer 6

my brother is here no problem now

Answer 7

see loga - logb=loga/b

Answer 8

let me post options :
a) 1/100 or 100
b) +or - 2
c) 10 or 1/10
d) 100
find the value of x

Answer 9

the answer should be a)

Answer 10

wrong dangol , how bhaiya: ?

Answer 11

x/sqrt(x) = sqrt(x) ... not x. My bad.

Answer 12

\[\log _{10}x-\log _{10}\sqrt{x}=\log _{10}x/\sqrt{x}\]
logx^1/2 = 2/ log x
(logx^1/2 )^2 =1
x=100

Answer 13

but it is saying that answer is a) , vishwesh bhaiya

Answer 14

it can be 1/100 also
log((1/100)^1/2)=-1
-1^2=1

Answer 15

oh thanks very much

Answer 16

Nice Work Vicky !!!

Answer 17

You really deserve a star

Answer 18

nice work both ( vishwesh and vicky bhaiya)

Answer 19

thanks by the way i am from India are you from tamilnadu

Answer 20

i am also from india ,i am from rajasthan , vicky bhaiya r u on facebook

Answer 21

No I am from Rajasthan

Answer 22

https://www.facebook.com/dsvivek007

Answer 23

i am also from india ,i am from rajasthan , vicky bhaiya r u on facebook

Answer 24

thanks

Answer 25

thanks bhaiya ;)

Answer 26

Please accept the request

Answer 27

i am also from india ,i am from rajasthan , vicky bhaiya r u on facebook

Answer 28

yes mine tooooo