How to solve exponential equations?
1. e^(2x) – 7e^(x) + 8 = 0 One of the rules of exponents is that when multiplying exponentials with the same base, you just add the exponents. According to that rule (e^x)(e^x) = e^(x+x) = e^2x So now your equation is the exponential form of the equation x²-7x+8. Use the quadratic equation to find value for e^x, then solve for x. a=1, b=-7, c=8 e^x = [-b ± √(b² – 4ac)]/2a = [ -7 ± √(49-32)]/2 = (-7±√17)/2 ——> exact value for e^x Take the natural log of e^x to solve for x ln(e^x) = ln((-7±√17)/2) x = ln((-7±√17)/2) (-7+√17)/2 ≈ -1.44 (-7-√17)/2 ≈ -5.56 Since both (-7+√17)/2 and (-7-√17)/2 are negative, the natural log of these numbers do not exist and there are no value of x that satisfy this equation NO SOLUTION. 2. 2^(0.
