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How do you solve this logarithmic equation 2logX equals log2 plus log3X-4? - Answers

assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2



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How do you solve this logarithmic equation 2logX equals log2 plus log3X-4? - Answers

https://math.answers.com/calculus/How_do_you_solve_this_logarithmic_equation_2logX_equals_log2_plus_log3X-4

assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2



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https://math.answers.com/calculus/How_do_you_solve_this_logarithmic_equation_2logX_equals_log2_plus_log3X-4

How do you solve this logarithmic equation 2logX equals log2 plus log3X-4? - Answers

assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2

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      assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2

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