Solve the equation (1 + x) to the 10th power minus 10.7347x equals 1.

Step-by-step solution

1. 1

   Move everything to one side and spot the free root. Define

   $f(x)=(1+x)^{10}-10.7347x-1$

   Then $f(0)=1-0-1=0$, so $x=0$ is an exact solution — no numerics needed. Equations of this shape (a growth factor raised to the number of periods, minus a linear payout) always have the trivial root, and the interesting one is the other crossing.

2. 2

   Differentiate to see the shape.

   $f'(x)=10(1+x)^9-10.7347,\qquad f''(x)=90(1+x)^8>0\ \ (x>-1)$

   Since $f''>0$, $f$ is strictly convex on $(-1,\infty)$, so it can cross zero at most twice. Also $f'(0)=10-10.7347=-0.7347<0$: the curve is still falling as it passes through the origin.

3. 3

   Locate the unique minimum. Solve $f'(x)=0$:

   $(1+x)^9=1.07347\quad\Rightarrow\quad x=1.07347^{1/9}-1=0.0079085$

   and the value there is $f(0.0079085)=-0.0029357$, genuinely below zero. (A frequently quoted figure of $-0.000291$ for this minimum is an order of magnitude too small.)

4. 4

   Count the roots. On $(-1,\,0.0079085)$ the function decreases from $f(-1^{+})=9.7347$ down to $-0.0029357$, giving exactly one crossing — that is $x=0$. On $(0.0079085,\infty)$ it increases from $-0.0029357$ to $+\infty$, giving exactly one more. So there are precisely two real roots and no negative ones ($f(-0.01)=+0.0117$ confirms the sign to the left of $0$).

5. 5

   Bracket and refine the second root. $f(0.015)=-0.00048$ and $f(0.016)=+0.00027$, so the root lies between them. Bisecting or applying Newton's method gives

   $x=0.01565489$

   to eight decimals — not $0.015845$, at which $f=+1.5\times 10^{-4}$, still clearly positive.

6. 6

   Verify by direct substitution. $(1.0156549)^{10}=1.16805066$, while $1+10.7347(0.0156549)=1.16805066$. The two agree to eight decimals, so the solution set is $x=0$ and $x\approx 0.0156549$.