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MISCELLANEOUS EXERCISE 6 (II) · Q179

Q.Solve: log⁡2x4+4log⁡42x=2\sqrt{\log_2 x^4}+4\log_4\sqrt{\dfrac2x}=2.

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Let t=log⁡2xt=\log_2x (need t≥0t\ge0 for the outer square root to be real, since log⁡2x4=4t\log_2x^4=4t).

log⁡2x4=4t=2t\sqrt{\log_2x^4}=\sqrt{4t}=2\sqrt{t}.

For the second term: log⁡42/x=12log⁡4(2/x)=12[log⁡42−log⁡4x]=12[12−t2]=1−t4\log_4\sqrt{2/x}=\dfrac12\log_4(2/x)=\dfrac12\left[\log_42-\log_4x\right]=\dfrac12\left[\dfrac12-\dfrac{t}{2}\right]=\dfrac{1-t}{4} (using log⁡42=12\log_42=\tfrac12 and log⁡4x=log⁡2x/log⁡24=t/2\log_4x=\log_2x/\log_24=t/2).

So 4log⁡42/x=1−t4\log_4\sqrt{2/x}=1-t.

The equation becomes 2t+(1−t)=2  ⟹  2t−t=12\sqrt{t}+(1-t)=2 \implies 2\sqrt{t}-t=1. …

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