Question

log abc^0.5+log abc =?

Answer 1

options are: 5,9,2,1

Answer 2

let me see if this what you wrote before we begin
\[\log abc^{\frac{1}{2}}+\log abc\]

Answer 3

is that right?

Answer 4

is it (abc)^0.5 or just ab(c^0.5) huge difference

Answer 5

so which is it?

Answer 6

hello!! you there?

Answer 7

yes the first one is ryt

Answer 8

will be
1/2log abc +log abc

Answer 9

so it is (abc)^1/2 yes

Answer 10

yes

Answer 11

okay then \(\log (abc)^{frac{1}{2}}+\log abc= \frac{1}{2} \log abc+\log abc\)
\(=\frac{3}{2}\log(abc)\)

Answer 12

that first line where frac12 supposed to be all to 1/2

Answer 13

so we get \(=\log(abc)^{\frac{3}{2}}\)

Answer 14

yea , next step

Answer 15

that is all what is it you want to achieve?

Answer 16

a different way of doing that is
\(\log (abc)^{\frac{1}{2}}+\log abc=\log \frac{(abc)^{\frac{1}{2}}}{abc}\)

Answer 17

but the options of this ques given are
a) 5
b) 9
c) 2
d) 1

which onw wud be the ans?

Answer 18

i meant \(=log[(abc)^0.5 (abc)]\) not division!!

Answer 19

\(=\log[(abc)^{0.5} (abc)]\)

Answer 20

if those are your options then you missing something in the question

Answer 21

alright

Answer 22

can you post the picture of your question

Answer 23

that must be the wrong ques. i pick this ques from mformaths.

Answer 24

bdw thankew for your help. :)

Answer 25

that question cannot be any of those number
no more simplification is required

Answer 26

yep

Answer 27

all you can do is something like this \(=3/2[\log a+\log b+ \log c]\)
which pretty much does not help for anything new

Answer 28

ok i have another question cud u help me in that?

Answer 29

post is in a different post though

Answer 30

okz