Question

if log (a+b/2)=1/2 (log a+log b) prove that a = b.

Answer 1

IF \[\log{(a+b/2)}=1/2 (\log{a}+\log{b})\] PROVE THAT A=B

Answer 2

firstly, how can you combine:\[\log(a)+\log(b)\]using the standard log rules?

Answer 3

LOG AB

Answer 4

correct.
next how can we take the 1/2 from outside the log to the inside, i.e:\[\frac{1}{2}\log(ab)\]using standard log rules?

Answer 5

\[\log{ab^1/2} \]

Answer 6

to be clearer, its actually:\[\log(ab)^\frac{1}{2}\]

Answer 7

log ab^1/2 or\[\log{\sqrt{ab}} \]

Answer 8

correct

Answer 9

so now we are left with:\[\log(\frac{a+b}{2})=\log(\sqrt{ab})\]

Answer 10

which means we can equate the things inside the logs

Answer 11

yes i tried but it didn't work out

Answer 12

we get:\[\frac{a+b}{2}=\sqrt{ab}\]correct?

Answer 13

yes

Answer 14

now?

Answer 15

so now square both sides

Answer 16

i did that it didn't work out could u tell

Answer 17

what do you get when you square both sides?

Answer 18

\[(a^2+2ab+b^2)/4=ab\]

Answer 19

now multiply both sides by 4 - what do you get?

Answer 20

\[a^2+2ab+b^2=4ab\]

Answer 21

correct - almost there now.
now subtract 4ab from both sides and what do you get?

Answer 22

\[a^2-2ab+b^2=0\]

Answer 23

perfect!

now can you see how to factorise \(a^2-2ab+b^2\)?

Answer 24

\[a^2-b^2=0\]

Answer 25

not quite, it should be:\[(a-b)^2=0\]

Answer 26

\[a^2=b^2\]
so a=b

Answer 27

your factorisation was incorrect

Answer 28

oh sorry

Answer 29

(a-b)^2

Answer 30

yup

Answer 31

now?

Answer 32

so you end up with:\[(a-b)^2=0\]which means?

Answer 33

if "something" squared is zero, then that "something" must be?

Answer 34

0

Answer 35

correct, so we get:\[(a-b)^2=0\implies a-b=0\]

Answer 36

since a-b=0 a=b

Answer 37

thats it - you have proved it!

Answer 38

actually i wrote it earlier but forgot to post :p

Answer 39

:)