Logarithm/ Indices HELP Urgent?

Question

2e^2x = 5e^(x+1) -2e^2

Find x

Answer 1

2e^2x = 2(e^x)²

5e^(x + 1) = 5e(e^x)

2e^2x = 5e^(x+1) -2e^2 becomes

2(e^x)² = 5e(e^x) - 2e²

Let e^x = t

2t² - 5et + 2e² = 0

(2t - e)(t - 2e) = 0

t = e/2 or 2e

e^x = e/2 or 2e

x = ln e/2 or ln2e

x = 1 - ln2 or 1 + ln2

Answer 2

First rearrange the equation....

=> 2e^2x-5e^(x+1)+2e^2=0
=> 2(e^x)^2 - 5e(e^x)+2e^2=0

The above equation is now of the form of a quadratic equation...ax^2+bx+c=0

and the solution to a quadratic equation is as follows:

x = ( (-b)+/- (b^2-4ac)^(1/2) ) /2a

Just substitute the values of a, b and c in to the above equation and calculate the two values of x.

Good luck

Answer 3

Let e^x = y

Then you equation takes the following form

2y^2 - 5ey + 2e^2 = 0

y = 2e , 0.5e

e^x = 2e , e^x = 0.5e

x = 1+ ln2

x = 1 - ln2

Answer 4

2e^2x can be written as 2(e^x)^2
5e^(x+1) can be written as 5*e*(e^x)
you replace e^x with a variable named y
you move everything on one side of the equation
you get 2y^2-5ey+2e^2=0 2nd grade equation? don't know name
delta is 25e^2-16e^2=9e^2
y1,2 is 1/4(5e+- sqrt(9e^2))=1/4(5e+-3e)= either 2e or e/2
now e^x=2e => x=ln(2e)=ln2 + 1
e^x=e/2 => x=ln(e/2)=1 - ln2

Answer 5

x= -1

Answer 6

2.e^x.e^x = 5e^x.e - 2e²
Let y = e^x
2y² = 5y.e - 2e²
2y² - 5e.y + 2e² = 0
(2y - e).(y - 2e) = 0
y = e/2, y = 2e
e^x = e/2 , e^x = 2e
x = ln (e/2) , x = ln (2e)
x = 0.307, x = 1.69

Answer 7

3^a=5^b=15^x this ability here: a circumstances log 3 = b circumstances log 5 = x circumstances log 15, so a = xlog15/ log 3 and b = xlog15/ log 5 Plug those into ab/a+b. you may then get 2 fractions multplied via one yet another, throughout 2 fractions added to a minimum of one yet another:(x log 15/ log 3)(x log 15/ log 5) divided via (x log 15/ log 3 + x log15/ log 5) the mathematics paintings you're able to tutor will look like this: Divide the numerator and denominator via x. so which you get (x)(log 15/ log 3)(log 15/ log 5) divided via (log 15/ log 3 + log 15/ log 5) The denominator would be rewritten as one fraction, it particularly is this: (log 15 circumstances log 5 + log 15 circumstances log 3) divided via log thrice log 5. Multiply the numerator, (x)(log 15/ log 3)(log 15/ log 5) via the turn-flop of this rewrite of the denominator. so which you have got (x)(log 15/ log 3)(log 15/ log 5) circumstances ( [log thrice log 5] divided via [log 15 circumstances log 5 + log 15 circumstances log 3] ) = (x)(log 15)(log 15) divided via (log 15 circumstances log 5 + log 15 circumstances log 3) Divide numerator and denominator via log 15. you may get x(log 15) divided via (log 5 + log 3). keep in mind that log p + log q = log (p circumstances q) so which you have got x(log 15) divided via ( log (5 circumstances 3) ) = x(log 15) divided via log 15, and this equals x; so this proves that ab/a+b = x .