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anonymous
 one year ago
i need someone to walk me through solving this equation:
200x + 1,000 = 10(1.5^x)
medal and fan.(:
anonymous
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.0Is that 1.5 to the x power?

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1lol i plugged this into my calculator and it took about a min to solve for an x value

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1well first i know we might aswell divide everything by 10 to get rid of it

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0first of all divide both sides by 10 so you can take ln both sides...........

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry it took so long, i had to deal with something else.

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1you can factor out a 20 on the right side of the equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have 200x + 100 = 1(1.5^x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why am i putting a twenty there?

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1ok so u didnt factor out the 10 from everything

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i divided 10 and 1000 by 10

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1should be 20x + 100 = 1.5^x

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1divide everything by 10...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i thought you only divide the sides that don't have x

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1what... u have confused me

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1just divide each value in the equation by 10 so 200/10 1000/10 and 10/10

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1200x/10 +1000/10 = 10/10(1.5^x) now solve that @grunge101

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i thought you weren't supposed to divide the numbers with x

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1why did u think that? remember that 200 in front of x is just a coefficient you still have to reduce it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434065203963:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0or first you can find common factor 200x+1000

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1@nelsonjedi made a good drawing of what you should do

freckles
 one year ago
Best ResponseYou've already chosen the best response.2in any case I think you are going to a numerical or graphing approach to approximate the solutions to the equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you @nelsonjedi

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.1im getting some strange values for x when i solve for x on my calc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so i'm a little it confused. sorry guys i keep having problems on my end . so first i divide everything by 10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have 20x + 10 = 1(1.5^x)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2what method are you suppose to use to approximate the solutions? Or are are you familiar with lambert function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0um, i don't really know. i just have to solve it.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and aren't you suppose to have: 200x + 1,000 = 10(1.5^x) this is equivalent to 20x+100=(1.5)^x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and why isn't the 1 in front of (1.5^x) ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[200x+1000=10(1.5)^x \\ \text{ divide both sides by 10} \\ \frac{200x+1000}{10}=\frac{10(1.5)^x}{10} \\ \frac{200x}{10}+\frac{1000}{10}=\frac{10}{10}(1.5)^x \\ 20x+100=1(1.5)^x \text{ you can put a 1 but \it is \not needed } \\ \text{ this is the same as } 20x+100=(1.5)^x \\ \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so if you want to solve this algebraically you will have to know the lambert function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, could you walk me through it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2are you sure you have to solve algebraically ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0honestly, 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
 one year ago
Best ResponseYou've already chosen the best response.0but 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
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.0but 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
 one year ago
Best ResponseYou've already chosen the best response.2\[20x+100=(1.5)^x \\ \text{ or } 20x+100=(\frac{3}{2})^x \\\] \[20(x+5)=(\frac{3}{2})^x \\ \text{ \let } t=x+5 \\ 20t=(\frac{3}{2})^{t5} \\ 20t(\frac{3}{2})^{5}=(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=\frac{1}{t}(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=t^{1}(\frac{2}{3})^{t} \\ \frac{1}{20}(\frac{2}{3})^5=t(\frac{2}{3})^t\\ \frac{1}{20}(\frac{2}{3})^5=t(e^{\ln(\frac{2}{3}) \cdot t}) \\ \ln(\frac{2}{3}) \frac{1}{20} (\frac{2}{3})^5=\ln(\frac{2}{3})te^{\ln(\frac{2}{3} )\cdot t} \\ \text{ take } W \text{ of both sides } \\ W(\frac{1}{20} \ln(\frac{2}{3}) (\frac{2}{3})^5)=\ln(\frac{2}{3}) \cdot t \\ \text{ so we have } t=\frac{1}{\ln(\frac{2}{3})} W(\frac{1}{20}\ln(\frac{2}{3}) (\frac{2}{3})^5) \] 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
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.0thank you! @freckles i just learned how to solve it approximately.
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