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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?

  • Hello, do you know why they would give you a question you can't solve in chapter 1? Also, what do you mean by index laws? Do you mean exponents? I am from US, I am assuming you are from the UK? – Some Guy Feb 12 '21 at 02:07
  • I think they are trying to see if I can apply multiple index (exponent) laws to solve the equation. –  Feb 12 '21 at 02:09
  • Solving this kind of equation is basically what logarithms do, so coming to it without context is frankly a little confusing. If you leverage the fact that the function $20 \times 10^{0.3t}$ is increasing, by trial and error, you can identify an interval that contains the answer. Some inputs will yield outputs that are too small, and others will yield outputs that are too large, and you can adjust the range until you close in on an interval that contains the answer, and perhaps even a narrow interval. But I don't know you can get a simple symbolic expression for it without logarithms. – leslie townes Feb 12 '21 at 02:12
  • Thank you, perhaps at this stage it is a trial and error process. –  Feb 12 '21 at 02:13
  • Can you use a calculator? – Some Guy Feb 12 '21 at 02:14
  • Check my answer, it uses calculator – Some Guy Feb 12 '21 at 02:19
  • Yes - calculator assumed –  Feb 12 '21 at 02:22

2 Answers2

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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.

Deepak
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$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 $6$.

Some Guy
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