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

6 years agoits the first one

6 years ago= 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 ago1/8=2^(-3)

6 years agooh yes - silly me

6 years agohow do we solve that?

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

6 years agomy equation has no real roots (as its -3)

6 years agoso 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 agoclearly no real solution for this

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

6 years agoyea

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

6 years agoyes - i was just going to tell u that

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

6 years ago*equate

6 years agoyes - thats what i did - but no real solutins

6 years agox^2 + 4^x = -3

6 years agoi could wolfram it i suppose (cheating!!)

6 years ago- yea two complex roots

6 years ago\[\huge 4^{2x}=(2^2)^{2x}=2^{2.2x}=2^{4x}\]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 agofinal answer \(x=-3\) and \(x=-1\)

6 years agobut isnt the original question 4^(2^x)? not 4^(2x)

6 years agothis ones driving me crazy!!!

6 years ago@mendeaar002 come here plz

6 years agohe cleared his last reply.. i think it was 4^(2x)

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

6 years agook - thats sorted then - thanx

6 years agosorry its 2^x^2(4^2x)=1/8

6 years ago