Answer:
x = 0.92
Step-by-step explanation:
To solve this, we need first use the exponent rule of [tex]e^{bc}=(e^b)^c[/tex] on [tex]e^{2x}[/tex]. We can break it down to [tex]e^{2x}=(e^x)^2[/tex]. We can now re-write as:
[tex]6(e^x)^{2}-13(e^x)-5=0[/tex]
This looks like a trinomial that we can middle term factorize by letting [tex]y=e^x[/tex]. Thus we can write and factorize and solve as shown below:
[tex]6(e^x)^{2}-13(e^x)-5=0\\6y^2-13y-5=0\\6y^2+2y-15y-5=0\\2y(3y+1)-5(3y+1)=0\\(2y-5)(3y+1)=0[/tex]
Thus, 2y-5 = 0 OR 3y+1 = 0
Solving we have y = 5/2 and y = -1/3
Now bringing back the original variable of letting y = e^x, we have:
1. 5/2 = e^x, and
2. -1/3 = e^x
Solving 1:
[tex]\frac{5}{2}=e^x\\ln(\frac{5}{2})=ln(e^x)\\x=ln(\frac{5}{2})[/tex]
Solving 2:
We will have x = ln (-1/3) WHICH IS NOT POSSIBLE because ln is never negative.
So our answer is x = ln (5/2)
Rounding to nearest hundredth: x = 0.92
