who is top please see 2=1 ??

Question

let a=b then
a^2 = ab ---------- (*a both)
a^2+(a^2-2ab) =ab+(a^2-2ab) - ------- ( +(a^2-2ab) both)
then take log base 10

log(2a^2-2ab) =log(a^2-ab)
log(2(a^2-ab)) = log(a^2-ab)
log2 + log(a^2-ab) = log(a^2-ab)
log 2=0
then
log2 = log1
that mean
2 = 1

where wrong!!!!!!???
a+a = a
mean 2a =a

let's
a=b
a^2 = ab ------(*a)
a^2-b^2 = ab-b^2 -----(-b^2)
then
2 ^(a+b)(a-b) = 2^ a(a-b)

(2^(a-b))^(a+b) =(2^(a-b)) ^a
mean
a+b =a (a =b)
a+a =a
2a = a !!!

Answer 1

since a = b
a^2 - ab = 0
try getting the log of that (whatever base)

log 0 = log (a^2-ab) is undefined, so there lies your problem
what you did was division by zero, "log style"

Answer 2

The problem is when you take the log of each side because the log function is only defined for positive values.

Since a^2 =ab,
(a^2- ab) =0
and, consequently you cannot find a log for it.

Answer 3

The problem here is that taking log (0) is an undefined operation. While you don't see this immediately, if you look closely...

log (a^2-ab) = log (ab-ab) = log (0)

Even though you subtract log (a^2-ab) from both sides, and it would seem like this would be legal, it is not because log (0) does not exist.

Answer 4

There is no real number x so that:

n^x = 0

And by definition any number^0 = 1.

Therefore

log(0) = undefined as well as ln(0)

Answer 5

log(0) is undefined. Go back and study a bit more ☺

Doug

Answer 6

This is a variation of a problem that's posted from time to time. The first time I saw it was in 1993.