How do you solve $9^{2 x + 1} = 12$ $9^(2x+1)=12$ ?

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Answer 1 · 2016-04-21T19:10:50

$x = \frac{1}{2} (\frac{ln 12}{ln 9} - 1) \approx 0.065$ $x=1/2 (ln12/ln9-1)~~0.065$

${ln 9}^{2 x + 1} = ln 12$ $ln 9^(2x+1) = ln12$

$(2 x + 1) ln 9 = ln 12$ $(2x+1)ln9=ln12$

$2 x + 1 = \frac{ln 12}{ln 9}$ $2x+1=ln12/ln9$

$2 x = \frac{ln 12}{ln 9} - 1$ $2x=ln12/ln9-1$

$x = \frac{1}{2} (\frac{ln 12}{ln 9} - 1) \approx 0.065$ $x=1/2 (ln12/ln9-1)~~0.065$

How do you solve 9^(2x+1)=1292x+1=129^(2x+1)=12?