Question

2^x^2(4^2^x)=1/8

Answer 1

its the first one

Answer 2

= 2^(x^2) * 2^(4^x) = 1/8
2^(x^2 + 4^x) = 2^3
x^2 + 4^x = 3

- does that make sense mukushla?

Answer 3

1/8=2^(-3)

Answer 4

oh yes - silly me

Answer 5

how do we solve that?

Answer 6

\[\large 4^{2^x}=2^{2^{x+1}}\]

Answer 7

my equation has no real roots (as its -3)

Answer 8

so we have
\[\huge 2^{x^2} (4^{2^x})=2^{x^2} 2^{2^{x+1}}=2^{(x^2+2^{x+1})}=\frac{1}{8}=2^{-3}\]so\[\huge x^2+2^{x+1}=-3\]

Answer 9

clearly no real solution for this

Answer 10

since \[x^2+2^{x+1} >0 \]for any real number

Answer 11

yea

Answer 12

oh man its \[\huge 2^{x^2} \times 4^{2x}=\frac{1}{8}\]

Answer 13

yes - i was just going to tell u that

Answer 14

well make the the bases equal (in this case 2) for both sides of equation... then euate the exponents

Answer 15

*equate

Answer 16

yes - thats what i did - but no real solutins

Answer 17

x^2 + 4^x = -3

Answer 18

i could wolfram it i suppose (cheating!!)

Answer 19

- yea two complex roots

Answer 20

\[\huge 4^{2x}=(2^2)^{2x}=2^{2.2x}=2^{4x}\]so
\[\huge 2^{x^2} \times 4^{2x}=2^{x^2} \times 2^{4x}=2^{x^2+4x}=\frac{1}{8}=2^{-3}\]\[\huge x^2+4x=-3\]solve for \(x\)

Answer 21

final answer \(x=-3\) and \(x=-1\)

Answer 22

but isnt the original question 4^(2^x)?

not 4^(2x)

Answer 23

this ones driving me crazy!!!

Answer 24

@mendeaar002

come here plz

Answer 25

he cleared his last reply.. i think it was 4^(2x)

Answer 26

@cwrw238

no problem with ur work...if 4^(2^x) there is no real solution

Answer 27

ok - thats sorted then - thanx

Answer 28

sorry its 2^x^2(4^2x)=1/8