how to solve this equation algebraically, show process please (e^(2x))-(4e^(X))-5=0
+33658 First let z = ex. Now the equation can be written as z2 - 4z - 5 = 0 This factorises into (z+1)(z-5)=0 so there are two solutions for z; namely z = -1 and z = 5. Because z is ex we have
ex = -1 and ex = 5
Now, for all real values of x, ex is positive, so there is no real solution to ex = -1
For the other possibility we take logs of both sides to get ln(ex) = ln(5) or just x = ln(5) (where ln is log to the base e).
If you want the ln(5) in approximate decimal form:
x=ln(5)=x=1.6094379124341004
+33658 First let z = ex. Now the equation can be written as z2 - 4z - 5 = 0 This factorises into (z+1)(z-5)=0 so there are two solutions for z; namely z = -1 and z = 5. Because z is ex we have
ex = -1 and ex = 5
Now, for all real values of x, ex is positive, so there is no real solution to ex = -1
For the other possibility we take logs of both sides to get ln(ex) = ln(5) or just x = ln(5) (where ln is log to the base e).
If you want the ln(5) in approximate decimal form:
x=ln(5)=x=1.6094379124341004
