Mathematics
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

its the first one

6 years ago
OpenStudy (cwrw238):

= 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?

6 years ago
OpenStudy (anonymous):

1/8=2^(-3)

6 years ago
OpenStudy (cwrw238):

oh yes - silly me

6 years ago
OpenStudy (cwrw238):

how do we solve that?

6 years ago
OpenStudy (anonymous):

$\large 4^{2^x}=2^{2^{x+1}}$

6 years ago
OpenStudy (cwrw238):

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

6 years ago
OpenStudy (anonymous):

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$\huge x^2+2^{x+1}=-3$

6 years ago
OpenStudy (anonymous):

clearly no real solution for this

6 years ago
OpenStudy (anonymous):

since $x^2+2^{x+1} >0$for any real number

6 years ago
OpenStudy (cwrw238):

yea

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (cwrw238):

yes - i was just going to tell u that

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

*equate

6 years ago
OpenStudy (cwrw238):

yes - thats what i did - but no real solutins

6 years ago
OpenStudy (cwrw238):

x^2 + 4^x = -3

6 years ago
OpenStudy (cwrw238):

i could wolfram it i suppose (cheating!!)

6 years ago
OpenStudy (cwrw238):

- yea two complex roots

6 years ago
OpenStudy (anonymous):

$\huge 4^{2x}=(2^2)^{2x}=2^{2.2x}=2^{4x}$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$$

6 years ago
OpenStudy (anonymous):

final answer $$x=-3$$ and $$x=-1$$

6 years ago
OpenStudy (cwrw238):

but isnt the original question 4^(2^x)? not 4^(2x)

6 years ago
OpenStudy (cwrw238):

this ones driving me crazy!!!

6 years ago
OpenStudy (anonymous):

@mendeaar002 come here plz

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

@cwrw238 no problem with ur work...if 4^(2^x) there is no real solution

6 years ago
OpenStudy (cwrw238):

ok - thats sorted then - thanx

6 years ago
OpenStudy (anonymous):

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

6 years ago