<s>[INST] Find the appropriate base $$213_x=139_{10}$$
$$21_x=1021_4$$
How would I solve this
 [/INST] $$213_x=139_{10}$$
$$2\cdot x^2+1\cdot x^1+3\cdot x^0=1\cdot 10^2+3\cdot 10^1+9\cdot10^0$$
$$2x^2+x+3=139$$
$$2x^2+x-136=0\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{1+8\cdot 136}}{4}=\frac{-1\pm33}{4}$$
$$x_1=8$$ the base $$x_2=-17/2$$
is not a correct base see comments below.
similarly
$$21_x=1021_4$$
$$2x+1=4^3+2\cdot 4 +1=73\Rightarrow x=36$$
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