Can anybody suggest me how I might solve the following equation?
$$(1+x)^{300} -125x(1+x)^{300} = 1$$
where x is the unknown that I want to solve for.
The actual question in full is
Q. Assume that your are going to retire in 25 years time. You want a mortgage of £100,000 now to extend and renovate your house but want to have it paid in full before you retire. The maximum repayment per month that your budge will allow id £800. What is the rate of interest you need from your bank to have the loan repaid in 300 monthly repayments (i.e 25 years).
So If the present value of each installment forms geometric sequence:
100,000 = 800/(1+i) + 800/(1+i)^2....
Where i is the monthly interest rate.
Then we can use the formula for Amortization - mortgages and loans:
A = Pi(1+i)^t/((1+i)^t -1)
Where :
A = repayment amount, P is the principle i is the term rate, and t is the term.
Then we get :
800 = 100,000i(1+i)^300/((1+i)^300-1)
Simplifying gives us:
$(1+i)^{300} -125i(1+i)^{300} = 1$
By the way the answer is 8.75% (yearly rate) which converts to a monthly rate of .70146% i.e a value of i of .0070146 which does indeed satisfy the equation above.
You seem to be making a mistake in setting up the equation to be solved. If the per-period discount rate is $v$, then we have $$100000=800\sum_{k=0}^{299}v^k=800\frac{v^{300}-1}{v-1}$$ or $$125(1-v)=1-v^{300},\tag1$$ assuming $v\neq1.$ If $x$ is the per-period interest rate, then $v = \frac 1{1+x}$,and we can rewrite $(1)$ as $$125x(1+x)^{299}=(1+x)^{300}-1$$ so it seems you made a slight error in setting up your equation, or it might be a difference in when the first payment is due. In my experience the first payment on a mortage is due when the loan is received. In any event, it will be easier to deal with in the form $(1)$. We rewrite $(1)$ as $$v=\frac{124+v^{300}}{125}\tag2$$ There's no way to solve $(2)$ explicitly, but we can solve it by iteration. Write $$f(v)=\frac{124+v^{300}}125$$ and pick some starting value for $v$, say $v=0$. Then $f(v)=f(0)=.992$. Take this as the new guess for $v$ and compute $f(v)$ again. $f(.992)=0.9927187727986595$. Continue in this manner until we find a $v$ such that $f(v)=v$ to sufficiently many decimals. We quickly arrive at $$v=0.9929610747842791$$ which gives $$x=0.007088822910052306$$ This is the monthly interest rate, so the nominal annual rate on the mortgage will be $12$ times that, or $$0.08506587492062767$$ I imagine that the expected answer is $8.5\%$.