AIME 2023 I · 第 2 题

Solution 1

Denote $x = \log_b n$. Hence, the system of equations given in the problem can be rewritten as

$$\begin{aligned} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{aligned}$$ Solving the system gives $x = 4$ and $b = \frac{5}{4}$. Therefore,

$$n = b^x = \frac{625}{256}.$$ Therefore, the answer is $625 + 256 = \boxed{881}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2

We can use the property that $\log(xy) = \log(x) + \log(y)$ on the first equation. We get $\log_b(bn) = 1 + \log_b(n)$. Then, subtracting $\log_b(n)$ from both sides, we get $(b-1) \log_b(n) = 1$, therefore $\log_b(n) = \frac{1}{b-1}$. Substituting that into our first equation, we get $\frac{1}{2b-2} = \sqrt{\frac{1}{b-1}}$. Squaring, reciprocating, and simplifying both sides, we get the quadratic $4b^2 - 9b + 5 = 0$. Solving for $b$, we get $\frac{5}{4}$ and $1$. Since the problem said that $b \neq 1$, $b = \frac{5}{4}$. To solve for $n$, we can use the property that $\log_b(n) = \frac{1}{b-1}$. $\log_\frac{5}{4}(n) = 4$, so $n = \frac{5^4}{4^4} = \frac{625}{256}$. Adding these together, we get $\boxed{881}$

~idk12345678

Solution 3 (quick)

There is some $x$ such that $n=b^{4x^2}$. The first equation becomes $2x=2x^2$, so $x=1$ since $x\ne 0$ following from the stated $b\ne 1$. Thus $n=b^{4\cdot 1^2}=b^4$, which turns the second equation into $4b=5$. Thus $b=\frac{5}{4}$ and $n= b^4=\frac{625}{256}$. Adding the numerator and denominator gives $\boxed{881}$.

Honestly, this problem is kinda well placed.

~Yrock

Video Solution by TheBeautyofMath

https://youtu.be/U96XHH23zhA

~IceMatrix

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=jxY7BBe-4gU