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How do you solve Log3 log2x - log(3x plus 7)? - Answers

To solve the equation ( \log_3(\log_2 x) - \log(3x + 7) = 0 ), first rewrite it as ( \log_3(\log_2 x) = \log(3x + 7) ). This implies ( \log_2 x = 3^{\log(3x + 7)} ). Next, convert ( \log(3x + 7) ) to base 3, and isolate ( x ) by converting back to exponential form. Finally, solve the resulting equation for ( x ).



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How do you solve Log3 log2x - log(3x plus 7)? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_Log3_log2x_-_log(3x_plus_7)

To solve the equation ( \log_3(\log_2 x) - \log(3x + 7) = 0 ), first rewrite it as ( \log_3(\log_2 x) = \log(3x + 7) ). This implies ( \log_2 x = 3^{\log(3x + 7)} ). Next, convert ( \log(3x + 7) ) to base 3, and isolate ( x ) by converting back to exponential form. Finally, solve the resulting equation for ( x ).



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_Log3_log2x_-_log(3x_plus_7)

How do you solve Log3 log2x - log(3x plus 7)? - Answers

To solve the equation ( \log_3(\log_2 x) - \log(3x + 7) = 0 ), first rewrite it as ( \log_3(\log_2 x) = \log(3x + 7) ). This implies ( \log_2 x = 3^{\log(3x + 7)} ). Next, convert ( \log(3x + 7) ) to base 3, and isolate ( x ) by converting back to exponential form. Finally, solve the resulting equation for ( x ).

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      To solve the equation ( \log_3(\log_2 x) - \log(3x + 7) = 0 ), first rewrite it as ( \log_3(\log_2 x) = \log(3x + 7) ). This implies ( \log_2 x = 3^{\log(3x + 7)} ). Next, convert ( \log(3x + 7) ) to base 3, and isolate ( x ) by converting back to exponential form. Finally, solve the resulting equation for ( x ).

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