Answer:
x = 0.92
Explanation:
To solve this, we need first use the exponent rule of
on
. We can break it down to
. We can now re-write as:
![6(e^x)^(2)-13(e^x)-5=0](https://img.qammunity.org/2020/formulas/mathematics/high-school/wa6knfo5m1xdrb2rp3xplh1scs6o5t8i8b.png)
This looks like a trinomial that we can middle term factorize by letting
. Thus we can write and factorize and solve as shown below:
![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](https://img.qammunity.org/2020/formulas/mathematics/high-school/9anq5zgk6xch8hrteo7w87q2971cio0tje.png)
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:
![(5)/(2)=e^x\\ln((5)/(2))=ln(e^x)\\x=ln((5)/(2))](https://img.qammunity.org/2020/formulas/mathematics/high-school/48qeskcrlq5f2qysl5cl71zb9ht68n17hv.png)
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