Solve the following equation:

$$
1.2065^{x}+1.2065^{x+1}+1.2065^{x+2}+1.2065^{x+3}=1.2065^{10}
$$
By factoring the right side and dividing by the known factors, after the calculations,

$$
1.2065^{x}=\frac{1.2065^{10}}{(1+1.2065)\left(1+1.2065^{2}\right)} \approx 1.206
$$

and thus $x \approx 1$.

Remark. The interesting aspect of the problem is that by making the sum of the exponents on the left side equal to the exponent on the right side, the correct solution is obtained!.[^0]


[^0]:    ${ }^{1}$ See W. Lietzmann: Wo steckt der Fehler? (Where is the error hidden?) with the following equation: $1.3247^{x}+1.3247^{x+1}+1.3247^{x+2}=$ $1.3247^{6}$ (B. G. Teubner, Leipzig, 1952, p. 83).