How do you solve $11^{x^{2}} = 25.4$ $11^(x^2) =25.4$ ?

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Answer 1 · 2016-03-17T10:34:05

$x = 1.1614$ $x=1.1614$ or $x = - 1.1614$ $x=-1.1614$

Taking log of both the sides in the equation $11^{x^{2}} = 25.4$ $11^(x^2)=25.4$ ,

$x^{2} \times log 11 = log 25.4$ $x^2xxlog11=log25.4$ or

$x^{2} = \frac{log 25.4}{log 11} = \frac{1.4048}{1.0414} = 1.349$ $x^2=(log25.4)/log11=1.4048/1.0414=1.349$

Hence $x = \pm 1.1614$ $x=+-1.1614$ i.e. $x = 1.1614$ $x=1.1614$ or $x = - 1.1614$ $x=-1.1614$

How do you solve 11^(x^2) =25.411x2=25.411^(x^2) =25.4?