1. Let's find the base $x$ for the number $(453)_x$ by understanding that digits represent coefficients for powers of $x$: $$4x^2+5x+3$$ The base must be higher than the largest digit, so $x > 5$. 2. For the equation $(45)_{10} = (63)_x$, rewrite both sides in decimal: $(45)_{10} = 45$ $(63)_x = 6x + 3$ Set equal: $$6x + 3 = 45$$ Solve for $x$: $$6x = 42 \\ x = 7$$ 3. For $(121)_x = (144)_8$, convert $(144)_8$ to decimal: $$1 imes 8^2 + 4 imes 8 + 4 = 64 + 32 + 4 = 100$$ Then express $(121)_x$ in decimal: $$1 imes x^2 + 2 imes x + 1$$ Set equal: $$x^2 + 2x + 1 = 100$$ Simplify: $$x^2 + 2x - 99 = 0$$ Solve the quadratic: $$x = \frac{-2 \pm \sqrt{4 + 396}}{2} = \frac{-2 \pm 20}{2}$$ Positive root: $$x = 9$$ 4. For $(145)_x = (x5)_{10}$, interpret $(x5)_{10}$ literally as the two-digit decimal number with digits $x$ and $5$: $$10x + 5$$ Express $(145)_x$ in decimal: $$1x^2 + 4x + 5$$ Set equal: $$1x^2 + 4x + 5 = 10x + 5$$ Simplify: $$x^2 + 4x + 5 - 10x - 5 = 0 \\ x^2 - 6x = 0$$ Factor: $$x(x - 6) = 0$$ Possible bases: $x = 0$ (not valid) or $x=6$ 5. For $34 + 56 = 112$, assume addition in some base $x$: Express each number in decimal: $$3x + 4$$ $$5x + 6$$ Sum: $$ (3x + 4) + (5x + 6) = 8x + 10$$ Right side is $112$ base $x$: $$1x^2 + 1x + 2 = x^2 + x + 2$$ Set equal: $$8x + 10 = x^2 + x + 2$$ Rearrange: $$x^2 + x + 2 - 8x - 10 = 0 \\ x^2 -7x -8 = 0$$ Solve quadratic: $$x = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2}$$ Positive roots: $$x = 8 ext{ or } x = -1$$ (discard negative) **Final answers:** - $(453)_x$: base $x > 5$ (depends on context) - $(45)_{10} = (63)_x$, $x = 7$ - $(121)_x = (144)_8$, $x = 9$ - $(145)_x = (x5)_{10}$, $x = 6$ - $34 + 56 = 112$ in base $x$, $x = 8$