F = P(1+i)^t

[Stricks]

When searching i I dont understand how F flips over the = sign. Why P is divided by F makes no sense to me
F = P(1+i)^t
147,500 = 120,500(1+i)⁴
(1+i)⁴ = 147,500/120,000 this step makes no sense to me, why does it happen?

Any feedback would be appreciated, thanks

 

[Cubist]

I think you made a typo with the number 120,X00 where digit X could be 0 or 5.

Let's use symbols, I'll show you the first few steps...

[math] F = P(1+i)^t [/math]
swap sides

[math] P(1+i)^t = F [/math]
in this step I show that "P" is not involved in the exponentiation (power) part of the left hand side

[math] P \times \left( (1+i)^t \right) = F [/math]
Can you continue from here? What do you need to do, to both sides, to eliminate the P from the left hand side?

 

[Stricks]

Thank you for your help, I managed to figure it out, dividing 200,000 from both sides, I have another problem if your interested in having a look haha. The fair side dice and the biased dice are both rolled so it would be 1/6 first time then 2/9 the second which is 3/18 and 4/18 that makes 7/18 which is an option but I think it's wrong, the probability of rolling 6 on the two dices can't be over a third

 

[Harry_the_cat]

Consider the simple case of say 3a=6.
If you divide both sides by 3, you get a=6/3 =2.
This is clearly correct since 3x2=6.
It's the same concept with your equation.

 

[JeffM]

Thank you for your help, I managed to figure it out, dividing 200,000 from both sides, I have another problem if your interested in having a look haha. The fair side dice and the biased dice are both rolled so it would be 1/6 first time then 2/9 the second which is 3/18 and 4/18 that makes 7/18 which is an option but I think it's wrong, the probability of rolling 6 on the two dices can't be over a third

Please put a new problem in a new thread.

What is the probability of rolling a 6 on the fair die?

What is the probability of rolling a 6 on the biased die?

P(A and B) = what?

What does it mean if two events are independent?