i need someone to walk me through solving this equation:
200x + 1,000 = 10(1.5^x)
medal and fan.(:

- anonymous

i need someone to walk me through solving this equation:
200x + 1,000 = 10(1.5^x)
medal and fan.(:

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- anonymous

Is that 1.5 to the x power?

- anonymous

yes

- sweetburger

lol i plugged this into my calculator and it took about a min to solve for an x value

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## More answers

- anonymous

haha

- sweetburger

well first i know we might aswell divide everything by 10 to get rid of it

- Nnesha

first of all divide both sides by 10 so you can take ln both sides.......-....

- anonymous

okay

- anonymous

one second please

- anonymous

okay i got it

- anonymous

sorry it took so long, i had to deal with something else.

- sweetburger

you can factor out a 20 on the right side of the equation

- anonymous

i have 200x + 100 = 1(1.5^x)

- anonymous

why am i putting a twenty there?

- sweetburger

ok so u didnt factor out the 10 from everything

- anonymous

yes i divided 10 and 1000 by 10

- sweetburger

should be 20x + 100 = 1.5^x

- anonymous

wait what

- sweetburger

divide everything by 10...

- anonymous

i thought you only divide the sides that don't have x

- anonymous

so 1000 and 10

- sweetburger

what... u have confused me

- sweetburger

just divide each value in the equation by 10 so 200/10 1000/10 and 10/10

- sweetburger

200x/10 +1000/10 = 10/10(1.5^x) now solve that @grunge101

- anonymous

i thought you weren't supposed to divide the numbers with x

- sweetburger

why did u think that? remember that 200 in front of x is just a coefficient you still have to reduce it

- anonymous

|dw:1434065203963:dw|

- Nnesha

or first you can find common factor 200x+1000

- sweetburger

@nelsonjedi made a good drawing of what you should do

- freckles

in any case I think you are going to a numerical or graphing approach to approximate the solutions to the equation

- anonymous

thank you @nelsonjedi

- sweetburger

im getting some strange values for x when i solve for x on my calc

- 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

yes right

- anonymous

okay, then what?

- anonymous

i have 20x + 10 = 1(1.5^x)

- freckles

what method are you suppose to use to approximate the solutions?
Or are are you familiar with lambert function?

- anonymous

um, i don't really know. i just have to solve it.

- anonymous

algebraically

- 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

oh yes, i meant 10.

- anonymous

and why isn't the 1 in front of (1.5^x) ?

- 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

okay i understand.

- freckles

so if you want to solve this algebraically you will have to know the lambert function

- anonymous

okay, could you walk me through it?

- freckles

are you sure you have to solve algebraically ?

- 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

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

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

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

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

\[20x+100=(1.5)^x \\ \text{ or } 20x+100=(\frac{3}{2})^x \\\]
\[20(x+5)=(\frac{3}{2})^x \\ \text{ \let } t=x+5 \\ 20t=(\frac{3}{2})^{t-5} \\ 20t(\frac{3}{2})^{5}=(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=\frac{1}{t}(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=t^{-1}(\frac{2}{3})^{-t} \\ \frac{1}{20}(\frac{2}{3})^5=t(\frac{2}{3})^t\\ \frac{1}{20}(\frac{2}{3})^5=t(e^{\ln(\frac{2}{3}) \cdot t}) \\ \ln(\frac{2}{3}) \frac{1}{20} (\frac{2}{3})^5=\ln(\frac{2}{3})te^{\ln(\frac{2}{3} )\cdot t} \\ \text{ take } W \text{ of both sides } \\ W(\frac{1}{20} \ln(\frac{2}{3}) (\frac{2}{3})^5)=\ln(\frac{2}{3}) \cdot t \\ \text{ so we have } t=\frac{1}{\ln(\frac{2}{3})} W(\frac{1}{20}\ln(\frac{2}{3}) (\frac{2}{3})^5) \]
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

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

thank you! @freckles i just learned how to solve it approximately.

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