NT prac 2

Note that $x^4_{10} = 777_b = 7(b^2 + b+1)_{10}$ . Therefore $7^4$ must be a divisor of $x^4$ . Assuming $x = 7$ , then

$\begin{aligned} b^2 + b + 1 & = 7^3 = 343 \\ b^2 + b - 342 & = 0 \\ (b-18)(b+19) & = 0 & \small \blue{\text{Since }b > 1} \\ \implies b & = 18 \end{aligned}$

Since we have assume $x^4 = 7^x$ , $\boxed{18}$ must be the smallest $b$ to satisfy the equation.

Turning $777$ into base $10,$ we get $7(b^2+b+1).$ Though the number is biquadratic, $7^3$ is the least value of $(b^2+ b+1).$

Solving the equation, we get $b=\boxed {18}$