# Thread: rearranging this formula

1. ## rearranging this formula

Hello,

I have just tried to put together a formula for savings which is attached below.

I have it for such a question
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I put in a lump sum amount of £3,500 into a new account. I also will be paying £900 of my wage per month into this account for 5 years. The gross interest rate will be 5%. How much will I end up with after this time?

where the answer is

γ = £3,500
α = 0.05
t = 5
β = £900

F=((γ)(exp^(αt)))+((β)((exp^(αt))-1)/(exp^(α/12))-1))

F = (( £3,500 )(exp^(0.05x5))) + (( £900 )(( exp^(0.05x5)) - 1)/( exp^(0.05/12)) -1 ))

F = (3500 x (1.284)) + ((900 x (((1.284) – 1)/((1.004)-1))

F = 4494.089 + (900 x (68.024))

F = 4494.089 + 61221.767

F = £65,715.86

Therefore I will have saved £65,715.86.

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Firstly is this formula correct? Because I have noticed in this calculator SAVINGS CALCULATOR ... I am getting a different figure. I have noticed that this Halifax calculator is missing the final months interest. So the question is, is my formula right? and if so can someone help me in re-arranging this formula, to solve for t and to solve for β as it is giving me a headache.

Thanks alot.

2. If the 5% compounds monthly (i = .05/12) and the 1st 900 deposit is
1 month after the initial deposit of 3500, then answer is 65,697.23

i = .05/12:

3500(1 + i)^60 = 4491.76

900[(1 + i)^60 - 1] / i = 61205.47 ; add 'em to get 65,697.23

3. excellent. thank you

now with this formula

F=((γ)(1+i)^12t) + ((β/i)((1 + i)^12t) - 1))

how do I rearrange this to solve for t?

4. ignore my last post mate.

thanks alot

5. Originally Posted by da`
F=((γ)(1+i)^12t) + ((β/i)((1 + i)^12t) - 1))
how do I rearrange this to solve for t?
Using this rule:
If a^p = x, then p = log(x) / log(a)