InterviewSolution
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InterviewSolution
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`3^(log_3log sqrtx) -log x + (log x)^2 - 3 =0` |
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Answer» `3^(log_3 logsqrtx) - logx +(logx)^2 - 3 = 0` We know, `a^(log_a b) = b`, so our equation becomes, `=>logsqrtx - logx +(logx)^2 - 3 = 0` `=>1/2logx - logx +(logx)^2 - 3 = 0` `=>(logx)^2 - logx/2 - 3 = 0` `=>2(logx)^2 - logx - 6 = 0` `=>2(logx)^2 - 4logx+3logx - 6 = 0` `=>(2logx+3)(logx - 2) = 0` `=>logx = -3/2 and logx = 2` `=>x = e^(-3/2) and x = e^2.` |
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