Question

solve for x..... e^2x=1

Answer 1

x=0

Answer 2

because anything to the power of 0 is 1

Answer 3

what???

Answer 4

\[b^0=1\] by definition

Answer 5

it isn't to the power of zero

Answer 6

you have
\[e^{2x}=1\] yes?

Answer 7

the equation is e to the power of 2x which equals 1

Answer 8

so i don't see how you get that x would be zero

Answer 9

right. so what power do you raise any number to to give 1?

Answer 10

ohhh zero?

Answer 11

yes so you know
\[2x=0\]

Answer 12

and therefore x = ...

Answer 13

ohhh okay thanks

Answer 14

yw

Answer 15

do you think i could askyou another question?

Answer 16

go ahead

Answer 17

ln(x^2)=ln(x+2)

Answer 18

solve for x

Answer 19

ok

Answer 20

\[\ln(x)\] is a "one to one" function which is a fancy way of saying that if
\[\ln(a)=\ln(b)\] then \[a=b\]

Answer 21

ohhh so x^2 equals x+2?

Answer 22

you are given that
\[\ln(x^2)=\ln(x+2)\] and the only way that is true is if
\[x^2=x+2\]

Answer 23

just be careful when you solve.

Answer 24

\[x^2-x-2=0\]
\[(x-2)(x+1)=0\]
\[x=2\] or \[x=-1\]

Answer 25

yes you are right. solve that

Answer 26

now we check the answers, because you cannot take the log of a negative number

Answer 27

but that is ok here because if x = -1 you get
\[\ln((-1)^2)=\ln(1)=0\] and also
\[\ln(-1+2)=\ln(1)=0\] so -1 works
also 2 works because
\[\ln(2^2)=\ln(2+2)=\ln(4)\]

Answer 28

ok so can i keep asking you questions or is it annoying?

Answer 29

one more then i have to go

Answer 30

cos^2x+2cosx=0

Answer 31

solve for x

Answer 32

factor to get
\[\cos(x)(\cos(x)+2)=0\]

Answer 33

first one give
\[\cos(x)=0\] so
\[x=\frac{\pi}{2}\] or \[x=\frac{3\pi}{2}\]

Answer 34

second one give
\[\cos(x)=-2\] which is not possible since cosine is between -1 and 1

Answer 35

thank you so much!

Answer 36

dont go!! haha

Answer 37

one more?