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I was wondering if it possible to correctly determine the solution for the following logarithmic equation to precision of one decimal point without a graphical calculator:

$$30\log_{10}(t)+t-72=0$$

Thank you

gordor
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  • Is this your homework ? –  Oct 25 '20 at 17:51
  • you can draw some pictures yourself. On paper. This increases with $t,$ and $t=10$ is too small. $t=100$ is too large. – Will Jagy Oct 25 '20 at 17:51
  • This equation cannot solved algebraically. So use an approximation method. Are you familiar with the Newton-Raphson method, for instance? – callculus42 Oct 25 '20 at 17:52
  • No, it's from a set of exercises for exam preparation; there no similar exercises in the textbook. I looked up for a way with Tylor expansion, by this is way beyond our level, also I don't know to deal with the error... – gordor Oct 25 '20 at 17:55
  • next $\sqrt{1000} \approx 31.62277$ gives $30 \cdot \frac{3}{2} + \sqrt{1000} = 45 + \sqrt{1000} \approx 76.622$ Suggest using rational $\delta$ with very small denominator and try $10^{1+ \delta}$ as well as $10^{2-\delta}$ In effect, move the problematic logarithm term to estimating $\sqrt[3]{10000}$ and similar. – Will Jagy Oct 25 '20 at 17:55
  • I am afraid that there is no easy answer by hand. –  Oct 25 '20 at 17:59
  • $ 30 x + 10^x = 72$ seems a bit better, with this $1 < x < 2$ – Will Jagy Oct 25 '20 at 18:11
  • Thank you, guys! – gordor Oct 25 '20 at 18:24
  • @WillJagy After observing that $10\sqrt{10}$ is close, rewrite the logarithm as $\log t = \log\frac{10\sqrt{10}}{10\sqrt{10}}t = \frac32+\log\frac t{10\sqrt{10}}$. From there, using the approximation $\log x \approx \frac2{\ln{10}}\frac{x-1}{x+1}$, which is good when $x\approx1$, we have $$30\left(\frac32+\frac2{\ln{10}}\frac{\frac t{10\sqrt{10}}-1}{\frac t{10\sqrt{10}}+1}\right)+t-72=0.$$ This is just a quadratic, and with the well-known approximation $\ln 10 \approx 2.3$ we get a very accurate $t\approx28.4$. – Théophile Feb 17 '23 at 20:29

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