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Solve the equation log3 (x + 17) - 2 = log3 2x
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\[ log_3 (x + 17) - 2 = log_3 2x\]\[ log_3 (x + 17) - 2log_33 = log_3 2x\]\[ log_3 (x + 17) - log_33^2 = log_3 2x\]\[ log_3 (\frac{x + 17 }{3^2} )= log_3 2x\]\[(\frac{x + 17 }{3^2} )= 2x\]I think you can do it from here
Things to note: (i) \(log_33 = 1\) (ii) \(n\log a = \log a^n\) (iii) \(\log a - \log b = \log \frac{a}{b}\) (iv) when you see \(log_n (x) = log_n (y)\), you can get x = y, where x and y are some polynomials (or monomials, it depends)
For (i), to generalise it, you may say \(log_nn=1\)
Note that it is for positive numbers. I am not sure of the negative one.
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