Question

4log2 (x2+1) = log2 .81​

Answer 1

$4 log_{2}( {x}^{2} + 1 ) = log_{2}(81)$

$log_{2}( {x}^{2} + 1 ) = \frac{ log_{2}(81) }{4}$

${x}^{2} + 1 = 3$

${x}^{2} = 3 - 1$

${x}^{2} = 2$

$x = +\sqrt{2}$