Can we find $a$ and $b$ as powers of $10$?

Can we find $a$ and $b$ as powers of $10$?

Question

Solution

Approach 1:

We have that $10a^2 = b^3$. Taking logs to base $10$ gives $\log_{10}10 + 2\log_{10}a=3\log_{10}b.$

But we know that $\log_{10} a = \log_{10}b + 1$, and so $1+2(\log_{10}b + 1)=3\log_{10}b.$

This gives $\log_{10}b = 3$, and so $b = 10^3$, and $a = 10^4$.

Approach 2:

We have

$$\begin{align*} \log_{10} a &= \log_{10} b+1\\ \implies \log_{10} a &= \log_{10} b + \log_{10} 10\\ \implies \log_{10} a &= \log_{10} (10b)\\ \implies a &= 10b. \end{align*}$$

Using the other equation, we find $100b^2 = \frac{b^3}{10},$ and so $b=0$ or $b=1000$.

However $\log_{10} 0$ is not defined, and hence $b=0$ is not a solution.

Thus $b=10^3$, which leads to $a=\pm 10^4$. Again, $\log_{10}a$ appears in the question so $a$ cannot be negative.

So we have $b=10^3$ and $a=10^4$.