e2x + 2ex = 8. First, let's define z to be ex Substituting gives us the following quadratic equation:
z2+2z=8 or z2+2z-8=0
Solving for z gives us two possible zs: 2 and minus 4.
As z is a power of e, only z=2 makes sense.
Back solving we get:
ex = 2
and the answer is x=ln(2). If you substitute in the answer, you get e2ln2+2eln2=8, or 4+4=8. ln(2) ~= 0.69315.
Zn(s) → Zn2+(aq) + 2e- and Fe3+(aq) + e- → Fe2+(aq)
It is one linear equation in two unknown variables: e and a.
It can be simplified to: 4f+2e
3lnx - ln2=4 lnx^3 - ln2=4 ln(x^3/2)=4 (x^3)/2=e^4 x^3=2e^4 x=[2e^4]^(1/3)
Simplified is 5e-2f.
2e - 12 = 7e + 8Add 12 to both sides: 2e = 7e + 20 subtract 7e from both sides: -5e = 20 divide both sides by -5: e = -4
Zn2+ + 2e- <--> Zno -0.7618 V
Mg equals Mg2+ plus 2e-
2e - 12 = 7e + 8 Subtract 2e from both sides: - 12 = 5e + 8 Subtract 8 from both sides: -20 = 5e Divide both sides by 5: -4 = e That is, e = -4
meant to be e^x = 2e^1-2x
9E-05x
E=mc^2A^2+B^2=C^2E/m=C^2=A^2+B^2E/m=A^2+B^2E=m(A^2+B^2)correct
Zn(s) → Zn2+(aq) + 2e- and Fe3+(aq) + e- → Fe2+(aq)
It is one linear equation in two unknown variables: e and a.
Mg equals Mg2+ plus 2e-
Ag+(aq) + e- → Ag(s) and Cu(aq) → Cu2+(aq) + 2e-
It can be simplified to: 4f+2e