I have come across the following equation in chapter 1 of a book that is testing applications of index laws. Logarithms are not covered until chapter 17. I can't figure out how to solve for $t$ without using logarithms.
$$ 1000 = 20 \times 10^{0.3t} $$
I have tried to find a common base and then solve for $t$.
Motivation: In what year will the population of Koala's exceed 1000?
On
$1000 < 20 \times 10^{0.3t} \implies 50 < 10^{0.3t} \implies 50 < (10^{0.3})^t. $ Now, we find the year such that $50 < (10^{0.3})^t$. $10^{0.3}$ is close to cube root of $10$ which is $2.15$, so we have $50 < (2.15)^t$. Now we can easily guess that $t=5$ doesn't solve this, but $t=6$ does, so in year $5$ is when population first exceeds $1000$ because the population reached $1000$ between years $5$ and $6$.
You cannot solve this "exactly" without logarithms. But maybe they want you to apply an approximation.
You may be expected to "know" that $2 \approx 10^{0.3}$
Applying that we get, $10^3 \approx 10^{0.3} \cdot 10 \cdot 10^{0.3t}$ which gets us to $1.3 + 0.3t \approx 3$ and $t \approx \frac {3-1.3}{0.3} = \frac{17}3$.
The result is within $0.06 \%$ of the "exact" value you get by taking logarithms.