Question

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${3}^{2x + 1} 4(3^{x} ) - 15 = 0$
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Answer 1

$\underline\bold \red{Thanks \: for \: your \: question} \\ \\ \bold \blue{Given : \: } \\ \\ {3}^{2x + 1} 4(3^{x} ) - 15 = 0 \\ \\ \bold \red{Solution :} \\ \\ = > {3}^{2x + 1} 4(3^{x} ) - 15 = 0 \\ \\ = > 3 {}^{2x + 1 + x} \times 4 - 15 = 0 \\ \\ = > 3 {}^{3x + 1} \times 4 - 15 = 0 \\ \\ = > 3 {}^{3x + 1} \times 4 = 15 \\ \\ = > 3 {}^{3x + 1} = \frac{15}{4} \\ \\ = > 3x + 1 = log_{3} \frac{15}{4 } \\ \\ = > 3x + 1 = \frac{ log \frac{15}{4} }{ log_{3} } \\ \\ = > 3x = \frac{ log \frac{15}{4} }{ log_{3} } - 1 \\ \\ = > x = \frac{ \frac{ log\frac{15}{4} }{ log_{3} } }{3} - \frac{1}{3} \\ \\ = > x = \frac{log\frac{15}{4} }{3 log3} - \frac{1}{3} \\ \\ => \bold\pink{ x = 0.242\: approx.}$

Answer 2

Answer:

Hope this will help you your answer will be

$x = 0.242approx$

Thank my all answers and follow