## anonymous one year ago Solve for x using the natural logarithm:

1. jdoe0001

kinda hard to do, when the expression is invisible

2. anonymous

$2\times4^{x} = 9.5e ^{-2x}$

3. anonymous

Thank you for your patience, @jdoe0001

4. anonymous

@SithsAndGiggles Can you help?

5. jdoe0001

hmmm can you pot a quick screenshot of the material?

6. anonymous

@zepdrix Can you help? I'm desperate :(

7. zepdrix

:D

8. zepdrix

Oh you left :c

9. anonymous

I'm back!

10. zepdrix

:D

11. anonymous

:D

12. zepdrix

$\large\rm 2\cdot4^{x} = 9.5\cdot e ^{-2x}$I don't like decimals, fractions are better, so I'm gonna do something a little sneaky. Let's umm.... let's multiply both sides by 2, then divide each side by 4,$\large\rm 4^x=\frac{19}{4}e^{-2x}$

13. zepdrix

Then take natural log each side I guess, ya? :)

14. zepdrix

$\large\rm \ln\left(4^x\right)=\ln\left(\frac{19}{4}e^{-2x}\right)$We have to apply a bunch of fun log rules to solve for x.

15. zepdrix

$\large\rm \color{orangered}{\log(a^b)=b\cdot \log(a)}$Do you see how this orange rule might help us on the left side of our equation?

16. anonymous

Ok!

17. anonymous

Give me one sec

18. anonymous

Wait why doesn't 4 cancel out when you divide it?

19. zepdrix

Umm if that was too confusing, we can just do it this way instead :)$\large\rm 2\cdot4^{x} = 9.5\cdot e ^{-2x}$Dividing by 2 gives us:$\large\rm 4^{x} = 4.75\cdot e ^{-2x}$I just didn't like the decimal value, so I changed it to a fraction, but we can leave it like this if it's easier to understand.

20. anonymous

Ok

21. zepdrix

So taking log of each side gives us$\large\rm \ln\left(4^x\right)=\ln\left(4.75\cdot e^{-2x}\right)$

22. anonymous

So you get x(ln4) = ln(4.75e^-2x) ?

23. anonymous

And now ln and e cancel each other out

24. zepdrix

Left side looks good. Woops! Don't try to cancel out the e just yet! Gotta do some more simplifying before you can do that.

25. zepdrix

$\large\rm \color{orangered}{x\ln4}=\ln\left(4.75\cdot e^{-2x}\right)$How bout the right side? How can we apply this blue rule?$\large\rm \color{royalblue}{\log(a\cdot b)=\log(a)+\log(b)}$

26. anonymous

x(ln4)= ln(4.75) + ln(e^-2x) ?

27. zepdrix

Ok great! Now we don't have the number in front of the e, so we can "cancel" them out like you wanted.

28. anonymous

xln4= ln4.75 - 2x ?

29. zepdrix

$\large\rm x \ln4=\ln4.75+\ln e^{-2x}$$\large\rm x \ln4=\ln4.75-2x$mmmm k good!

30. zepdrix

Let's get all of our x's to one side and try some factoring.

31. anonymous

x(ln4) + 2x= ln4.75 ?

32. zepdrix

cool :) the terms on the left side both have something in common. try to factor it out :O

33. anonymous

x(ln4 + 2)= ln4.75 ?

34. zepdrix

Good good good, how you gonna wrap it up? :)

35. anonymous

x = 0.4601 ?

36. zepdrix

Yayyy good job \c:/ If your teacher wants you to leave it as an exact value you would have:$\large\rm x=\frac{\ln4.75}{2+\ln4}$ But yes, that's a correct decimal approximation.

37. anonymous

Yay!!! You are the BEST!!! Thanks so much!