anonymous
  • anonymous
Solve for x using the natural logarithm:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jdoe0001
  • jdoe0001
kinda hard to do, when the expression is invisible
anonymous
  • anonymous
\[2\times4^{x} = 9.5e ^{-2x}\]
anonymous
  • anonymous
Thank you for your patience, @jdoe0001

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anonymous
  • anonymous
@SithsAndGiggles Can you help?
jdoe0001
  • jdoe0001
hmmm can you pot a quick screenshot of the material?
anonymous
  • anonymous
@zepdrix Can you help? I'm desperate :(
zepdrix
  • zepdrix
:D
zepdrix
  • zepdrix
Oh you left :c
anonymous
  • anonymous
I'm back!
zepdrix
  • zepdrix
:D
anonymous
  • anonymous
:D
zepdrix
  • 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,\[\large\rm 4^x=\frac{19}{4}e^{-2x}\]
zepdrix
  • zepdrix
Then take natural log each side I guess, ya? :)
zepdrix
  • 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.
zepdrix
  • 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?
anonymous
  • anonymous
Ok!
anonymous
  • anonymous
Give me one sec
anonymous
  • anonymous
Wait why doesn't 4 cancel out when you divide it?
zepdrix
  • 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.
anonymous
  • anonymous
Ok
zepdrix
  • zepdrix
So taking log of each side gives us\[\large\rm \ln\left(4^x\right)=\ln\left(4.75\cdot e^{-2x}\right)\]
anonymous
  • anonymous
So you get x(ln4) = ln(4.75e^-2x) ?
anonymous
  • anonymous
And now ln and e cancel each other out
zepdrix
  • 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.
zepdrix
  • 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?\[\large\rm \color{royalblue}{\log(a\cdot b)=\log(a)+\log(b)}\]
anonymous
  • anonymous
x(ln4)= ln(4.75) + ln(e^-2x) ?
zepdrix
  • zepdrix
Ok great! Now we don't have the number in front of the e, so we can "cancel" them out like you wanted.
anonymous
  • anonymous
xln4= ln4.75 - 2x ?
zepdrix
  • zepdrix
\[\large\rm x \ln4=\ln4.75+\ln e^{-2x}\]\[\large\rm x \ln4=\ln4.75-2x\]mmmm k good!
zepdrix
  • zepdrix
Let's get all of our x's to one side and try some factoring.
anonymous
  • anonymous
x(ln4) + 2x= ln4.75 ?
zepdrix
  • zepdrix
cool :) the terms on the left side both have something in common. try to factor it out :O
anonymous
  • anonymous
x(ln4 + 2)= ln4.75 ?
zepdrix
  • zepdrix
Good good good, how you gonna wrap it up? :)
anonymous
  • anonymous
x = 0.4601 ?
zepdrix
  • 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.
anonymous
  • anonymous
Yay!!! You are the BEST!!! Thanks so much!

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