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How do I solve this log equation? I think it's impossible to solve this. Correct me of I'm wrong.

$$ 8\log_{10}\left(\dfrac {50-t}{45-t}\right)=5\log_{10}\left(\dfrac {50-t}{40-t}\right)$$

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closed as off-topic by Namaste, Davide Giraudo, zoli, Henrik, C. Falcon Apr 2 '17 at 23:16

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  • $\begingroup$ The Base is of log is 10 $\endgroup$ – Aditya DS Apr 2 '17 at 19:28
  • $\begingroup$ @projectilemotion It's irrelevant. $\endgroup$ – user228113 Apr 2 '17 at 19:29
  • $\begingroup$ @G.Sassatelli why? $\endgroup$ – Aditya DS Apr 2 '17 at 19:30
  • $\begingroup$ How can I solve this? $\endgroup$ – Aditya DS Apr 2 '17 at 19:30
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    $\begingroup$ @AdityaDS The logarithm in one base is a constant times the logarithm in an other base. So if you change the base of the logarithm on both sides, you multiply both sides by the same number. $\endgroup$ – Paul Apr 2 '17 at 19:37
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You can move the coefficients into the log to get $$\log \left(\frac{50-t}{45-t}\right)^8 = \log \left(\frac{50-t}{40-t}\right)^5.$$ (Note that G. Sassatelli in the comment below is right that this might introduce extra spurious solutions, since solutions to the new equation where $\frac{50-t}{45-t}$ is negative are not solutions of the original equation. Therefore any solution to the new equation with $45\leq t \leq 50$ should be discarded at the end of the calculation.)

You can then exponentiate both sides, eliminate the common numerator, and rearrange to get the polynomial equation $$(40-t)^5(50-t)^3 - (45-t)^8 = 0.$$ Things don't look too promising from here, but you can use e.g. Wolfram Alpha to approximate a solution $t\approx 55.431.$

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  • $\begingroup$ If we are to solve it numerically at the end, then the initial equation is numerically more stable compared to powers of $8$ in this one. This appear to be progress, but ironically having the $8$ and $5$ outside the log help for a better convergence. Sometimes manual solving and numerical solving are antagonist. :p $\endgroup$ – zwim Apr 2 '17 at 21:56
  • $\begingroup$ @zwim Fair point. I guess the purpose of my answer is to show how one would go about trying to solve the equation exactly, in order to conclude that it is "impossible." Once that conclusion has been reached you can go back and analyze the best way of numerically solving the original problem. $\endgroup$ – user7530 Apr 3 '17 at 7:12
  • $\begingroup$ I agree, I would have done the same, thus the smiley at the end. $\endgroup$ – zwim Apr 3 '17 at 7:31

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