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  • Solve the equation  <img alt="\\\mathrm{2\log_{\;3}x+\log_{\;3}(x^2-3) = \log_{3}0.5 + 5^{\log_{\;5}(\log_{\;3}8)}}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5C%5C%5Cmathrm%7B

Solve the equation  \\\mathrm{2\log_{\;3}x+\log_{\;3}(x^2-3) = \log_{3}0.5 + 5^{\log_{\;5}(\log_{\;3}8)}}

Option: 1

x=0


Option: 2

x=-2


Option: 3

x=2


Option: 4

none of the above


Answers (1)

best_answer

Using properties of logarithm, this equation can be written as 

\\\mathrm{\log_{\;3}x^2+\log_{\;3}(x^2-3) = \log_{3}0.5 + \log_{\;3}8} \\\\ log_3 (x^2.(x^2-3)) = log_3(0.5*8)\\\\\text{Now we have same base of log on both sides, so log can be removed from both sides}\\\\\mathrm{x^2(x^2-3)=0.5\cdot 8} \\\mathrm{x^2(x^2-3) = 4}

\\x^4 - 3x^2 - 4 = 0\\ Let\,\,x^2\,=\,t\\ t^2 - 3t - 4 = 0\\ (t-4)(t+1) =0\\ t = 4\,\,or\,\,t = -1\\ x^2 = 4\,\,or\,\,x^2 = -1\\ x = \pm2

Now check whether  x= 2 and x = -2 lies in the domain of original equation

For x = -2 first term in the equation is not defined, so it is rejected

But for x = 2, all the terms are defined

So x = 2 is the only answer

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