Exponent Equals to the Number of Digits

$d=\lfloor \log_{10} x^d \rfloor +1 = \lfloor d \cdot \log_{10} x \rfloor +1 \implies \lfloor d \cdot \log_{10} x \rfloor =d-1$

we definitely need $x<10$ , otherwise $\lfloor d \cdot \log_{10} x \rfloor \geq d$

then, we need

$d-1 \leq d\cdot \log_{10} x \implies \frac{d-1}{d} \leq \log_{10} x \implies 1-\frac{1}{d} \leq \log_{10} x$

$\implies -\frac{1}{d} \leq \log_{10} x-1 = \log_{10} \frac{x}{10} \implies d \leq \frac{-1}{\log_{10} \frac{x}{10}}$

now, notice $1<x<10$ and the higher $x$ is the higher the upper limit of $d$ can be. So, JUST take $x=9$ , which gives $d \leq 21.85$ . So, JUST take $d=21$ . You know, just do it!

We can set up inequalities if we make an example.

If $d$ is equal to $3$ , then $x^{d}$ could be $100,$ $101,$ $102, ...,$ or $999$ (numbers that have $3$ digits). Therefore, it must be between $99$ and $1000$ . If $d=4$ , then $x^{d}$ could be $1000,$ $1001,$ $1002,$ $...,$ or $9999$ (numbers that have $4$ digits). Therefore, it must be between $999$ and $10000$ .

Through those examples, we can deduce that

$10^{d-1}-1 < x^{d} < 10^{d}$

Since $x^{d} < 10^{d}$ and we're looking for the largest value, $x$ must be equal to $9$ because it is the largest integer before $10$ .

$\begin{aligned} & \log_9 (10^{d-1}-1) < d < \log_9 (10^{d}) & \\ \implies & \log_9 (10^{d-1}-1) < d < d \log_9 10 & \space \space \small \color{#3D99F6}\text{(divide all expressions by} \space d \text{)} \\ \implies & \frac{\log_9 (10^{d-1}-1)}{d} < 1 < \log_9 10 & \end{aligned}$

In the left expression, we find that the largest integer value for $d$ such that it would equate less than $1$ is $21$ .

$\frac{\log_9 (10^{21-1}-1)}{21} = 0.998... < 1$

$\begin{array}{cc} x^{d} = 9^{21} & \implies 9×21 = \boxed{189} \end{array}$