anonymous
  • anonymous
i need someone to walk me through solving this equation: 200x + 1,000 = 10(1.5^x) medal and fan.(:
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Is that 1.5 to the x power?
anonymous
  • anonymous
yes
sweetburger
  • sweetburger
lol i plugged this into my calculator and it took about a min to solve for an x value

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
haha
sweetburger
  • sweetburger
well first i know we might aswell divide everything by 10 to get rid of it
Nnesha
  • Nnesha
first of all divide both sides by 10 so you can take ln both sides.......-....
anonymous
  • anonymous
okay
anonymous
  • anonymous
one second please
anonymous
  • anonymous
okay i got it
anonymous
  • anonymous
sorry it took so long, i had to deal with something else.
sweetburger
  • sweetburger
you can factor out a 20 on the right side of the equation
anonymous
  • anonymous
i have 200x + 100 = 1(1.5^x)
anonymous
  • anonymous
why am i putting a twenty there?
sweetburger
  • sweetburger
ok so u didnt factor out the 10 from everything
anonymous
  • anonymous
yes i divided 10 and 1000 by 10
sweetburger
  • sweetburger
should be 20x + 100 = 1.5^x
anonymous
  • anonymous
wait what
sweetburger
  • sweetburger
divide everything by 10...
anonymous
  • anonymous
i thought you only divide the sides that don't have x
anonymous
  • anonymous
so 1000 and 10
sweetburger
  • sweetburger
what... u have confused me
sweetburger
  • sweetburger
just divide each value in the equation by 10 so 200/10 1000/10 and 10/10
sweetburger
  • sweetburger
200x/10 +1000/10 = 10/10(1.5^x) now solve that @grunge101
anonymous
  • anonymous
i thought you weren't supposed to divide the numbers with x
sweetburger
  • sweetburger
why did u think that? remember that 200 in front of x is just a coefficient you still have to reduce it
anonymous
  • anonymous
|dw:1434065203963:dw|
Nnesha
  • Nnesha
or first you can find common factor 200x+1000
sweetburger
  • sweetburger
@nelsonjedi made a good drawing of what you should do
freckles
  • freckles
in any case I think you are going to a numerical or graphing approach to approximate the solutions to the equation
anonymous
  • anonymous
thank you @nelsonjedi
sweetburger
  • sweetburger
im getting some strange values for x when i solve for x on my calc
anonymous
  • anonymous
okay so i'm a little it confused. sorry guys i keep having problems on my end . so first i divide everything by 10
Nnesha
  • Nnesha
yes right
anonymous
  • anonymous
okay, then what?
anonymous
  • anonymous
i have 20x + 10 = 1(1.5^x)
freckles
  • freckles
what method are you suppose to use to approximate the solutions? Or are are you familiar with lambert function?
anonymous
  • anonymous
um, i don't really know. i just have to solve it.
anonymous
  • anonymous
algebraically
freckles
  • freckles
and aren't you suppose to have: 200x + 1,000 = 10(1.5^x) this is equivalent to 20x+100=(1.5)^x
anonymous
  • anonymous
oh yes, i meant 10.
anonymous
  • anonymous
and why isn't the 1 in front of (1.5^x) ?
freckles
  • freckles
\[200x+1000=10(1.5)^x \\ \text{ divide both sides by 10} \\ \frac{200x+1000}{10}=\frac{10(1.5)^x}{10} \\ \frac{200x}{10}+\frac{1000}{10}=\frac{10}{10}(1.5)^x \\ 20x+100=1(1.5)^x \text{ you can put a 1 but \it is \not needed } \\ \text{ this is the same as } 20x+100=(1.5)^x \\ \]
anonymous
  • anonymous
okay i understand.
freckles
  • freckles
so if you want to solve this algebraically you will have to know the lambert function
anonymous
  • anonymous
okay, could you walk me through it?
freckles
  • freckles
are you sure you have to solve algebraically ?
anonymous
  • anonymous
honestly, i have no idea. this is something i learned a long time ago, and i struggled with it then. i don't remember what the process to solve it was called.
anonymous
  • anonymous
but i need to just solve this problem so i can understand how to do it and move on to the further step of this sort of equation using systems of equations approximately.
freckles
  • freckles
The equation you want to solve currently is done in most algebra classes by finding approximations to the solutions using numerical approach or graphing approach. I would graph both expressions (the left and right of the equal sign expressions) and see where they intersect. The intersection of these curves (line) are the solutions.
anonymous
  • anonymous
well i don't want to solve it using approximation and systems of equations and whatnot. right now, i need to solve it a step backwards.
anonymous
  • anonymous
but you know what, i don't really need help anymore. it doesn't matter. i'll figure it out later. you'll still get a medal for trying so hard and dedicating so much of your time tonight to helping me. thank you. :)
freckles
  • freckles
\[20x+100=(1.5)^x \\ \text{ or } 20x+100=(\frac{3}{2})^x \\\] \[20(x+5)=(\frac{3}{2})^x \\ \text{ \let } t=x+5 \\ 20t=(\frac{3}{2})^{t-5} \\ 20t(\frac{3}{2})^{5}=(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=\frac{1}{t}(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=t^{-1}(\frac{2}{3})^{-t} \\ \frac{1}{20}(\frac{2}{3})^5=t(\frac{2}{3})^t\\ \frac{1}{20}(\frac{2}{3})^5=t(e^{\ln(\frac{2}{3}) \cdot t}) \\ \ln(\frac{2}{3}) \frac{1}{20} (\frac{2}{3})^5=\ln(\frac{2}{3})te^{\ln(\frac{2}{3} )\cdot t} \\ \text{ take } W \text{ of both sides } \\ W(\frac{1}{20} \ln(\frac{2}{3}) (\frac{2}{3})^5)=\ln(\frac{2}{3}) \cdot t \\ \text{ so we have } t=\frac{1}{\ln(\frac{2}{3})} W(\frac{1}{20}\ln(\frac{2}{3}) (\frac{2}{3})^5) \] but recall t=x+5 \[x=\frac{1}{\ln(\frac{2}{3})}W(\frac{1}{20}\ln(\frac{2}{3})(\frac{2}{3})^5)-5\]
freckles
  • freckles
I post this thinking this is not what you really want and you are looking for an approach that is more likely to occur in algebra class which for this type of equation is a most likely some numerical/graphing type approach. But the above is I would consider very algebraic approach.
anonymous
  • anonymous
thank you! @freckles i just learned how to solve it approximately.

Looking for something else?

Not the answer you are looking for? Search for more explanations.