Question

Interesting question , still thinking

Answer 1

\[\huge 2^{2x ^{2}}+2^{x ^{2}+2x+2}=2^{5+4x}\]
Find x

Answer 2

Solve for x , I mean

Answer 3

\[\large 2^{2x^2-4x-5} + 2^{x^2-2x-3} -1 = 0\]
\[\large 2^{2(x^2-2x-2)-1} + 2^{x^2-2x-2-1} -1 = 0\]

Answer 4

\[\large 2^{2(x^2-2x-2)} + 2^{x^2-2x-2} -2 = 0\]

Answer 5

you knw what to do next ?:)

Answer 6

wait wait reading your reply

Answer 7

what did u do , did u transfer the R.h.s power to L.hS

Answer 8

divide both sides by the RHS term

Answer 9

oh wait

Answer 10

It would be -2 or 0 ?

Answer 11

those solutions are for what ?

Answer 12

no no no ,
the constant term in the expression above i am asking about

Answer 13

the quadratic would be \[\large u^2 + u-2 = 0\]

right ?

Answer 14

\[\large 2^{2(x^2-2x-2)} + 2^{x^2-2x-2} -2 = 0\]
or
\[\large 2^{2(x^2-2x-2)} + 2^{x^2-2x-2} = 0\]

Answer 15

whats a/a = ?

Answer 16

1 or 0 ?

Answer 17

1

Answer 18

then ? the RHS will be 1 after dividing it by itself, eh ?

Answer 19

no no taht

Answer 20

\[\large 2^{2(x^2-2x-2)-1} + 2^{x^2-2x-2-1} -1 = 0\]
After this what u did

Answer 21

multiply throught the equation by "2^1"

Answer 22

\[\large \color{red}{2^1}2^{2(x^2-2x-2)-1} + \color{red}{2^1}2^{x^2-2x-2-1} -\color{red}{2^1}1 = \color{red}{2^1}0\]

Answer 23

fine ,
then quadratic
u^2+u-2

Answer 24

u^2+u-2 = 0

solve it

Answer 25

yes waigt

Answer 26

[-1+/- 3 ]/2

Answer 27

is that u or x ?

Answer 28

u

Answer 29

that quadratic is factorable

Answer 30

u^2+u-2 = 0

(u+2)(u-1) = 0

Answer 31

my bad , yes sry

Answer 32

one of the values of u makes no sense, which one is it ?

Answer 33

can \(\large 2^{something ~real}\) can ever be negative ?

Answer 34

nope

Answer 35

that leaves you with u = 1

Answer 36

-2 not accepted

Answer 37

yes, when does an exponential become 1 ?

Answer 38

a^0

Answer 39

^^