I need a bit of help rearranging this equation

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In summary, the equation y = 50 (5 - 0.67^x) is being rearranged to isolate for x. The process involves dividing both sides by 50, subtracting 5, and taking the logarithm of both sides. The solution for x is x = [log((250-y)/50)/log(.67).
  • #1
Hfuais
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1. The equation is : y = 50 ( 5 - 0.67 ^ x)



2. Just basic rearranging, I need to isolate x.



3. So far I have..
y = 50 ( 5 - 0.67 ^ x)
y - 50=( 5 - 0.67 ^ x)
y - 55 = - 0.67 ^ x)
Now is the bit where I am stuck, seeing it is raised as a power? I know it's simple, I just can't remember! :/
 
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  • #2
Remember what the natural log (ln) is and this becomes simple.

[itex]55 - y = .67^x[/itex]

[itex]ln(55 - y) = ln(.67^x)[/itex]

[itex]ln(55 - y) = xln(.67)[/itex]

[itex]x = ln(55 - y) / ln(.67)[/itex]

Is this what you needed?
 
  • #3
You are doing pretty much everything wrong. To begin with, you do not just "move to the other side and change the sign". I wish that phrase we banned!

To "solve for x" or "make x the subject" (the first is American English, the second British English) you need to undo everything that is done to x, by doing the opposite, on both sides.

The first thing you want to eliminate is that "50" and it multiplies the rest of the right side and the opposite of multiplying is dividing. Divide both sides by 50 to get [itex]y/50= 5- 0.67^x[/itex]. Now, we have 5 added on the right so subtract 5 from both sides: [itex]y/50- 5= -0.67^x[/itex]. That "-" indicates a multiplication by -1 so divide by -1 (which, yes, is the same as multiplying by -1): [itex]5- y/50= 0.67^x[/itex]. Note that 5- y/50 is NOT (5- y)/50. We could write 5= 250/50 so that 5- y/50= (250- y)/50 so that [itex]5- y/50= (250- y)/50= 0.67^x[/itex].

Now the right side is a exponential function- the "x" is in the exponent. And the opposite of an exponential is a logarithm. That is, take the logarithm of both sides:
log((250- y)/50)= log(.67^x)= xlog(.67)

Finally, since we now have the right side reduced to x times something, divide by it: x= [log((250- y)/50)/log(.67).
 

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1. How do I rearrange an equation?

Rearranging an equation involves isolating the variable that you are trying to solve for on one side of the equation. This can be done by using algebraic operations such as addition, subtraction, multiplication, and division to move terms from one side of the equation to the other.

2. What is the purpose of rearranging an equation?

Rearranging an equation allows you to solve for a specific variable or isolate a certain term in the equation. This can help in problem solving and understanding the relationship between different variables and quantities in a given situation.

3. How do I know if I have rearranged the equation correctly?

You can check if you have rearranged the equation correctly by plugging in the values of the variables into the original equation and making sure that both sides of the equation are equal. If they are equal, then you have rearranged the equation correctly.

4. Can I rearrange any type of equation?

Yes, you can rearrange any type of equation as long as you use algebraic operations and follow the rules of algebra. However, some equations may be more complex and require more steps to rearrange.

5. Are there any tips for rearranging equations?

One helpful tip for rearranging equations is to always start by simplifying both sides of the equation before attempting to rearrange. This can make the process easier and help avoid making mistakes. It is also important to keep track of the algebraic operations used and make sure they are applied correctly on both sides of the equation.

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