Question

Solve for x using the natural logarithm:

Answer 1

kinda hard to do, when the expression is invisible

Answer 2

\[2\times4^{x} = 9.5e ^{-2x}\]

Answer 3

Thank you for your patience, @jdoe0001

Answer 4

@SithsAndGiggles Can you help?

Answer 5

hmmm can you pot a quick screenshot of the material?

Answer 6

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

Answer 7

:D

Answer 8

Oh you left :c

Answer 9

I'm back!

Answer 10

:D

Answer 11

:D

Answer 12

\[\large\rm 2\cdot4^{x} = 9.5\cdot e ^{-2x}\]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,\[\large\rm 4^x=\frac{19}{4}e^{-2x}\]

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Ok!

Answer 17

Give me one sec

Answer 18

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

Answer 19

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.

Answer 20

Ok

Answer 21

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

Answer 22

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

Answer 23

And now ln and e cancel each other out

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

xln4= ln4.75 - 2x ?

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

x(ln4 + 2)= ln4.75 ?

Answer 34

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

Answer 35

x = 0.4601 ?

Answer 36

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.

Answer 37

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