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Exercise 2.12 · Q3

Q.Solve log⁡8x+log⁡4x+log⁡2x=11\log_8 x+\log_4 x+\log_2 x=11.

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Step 1. log⁡8x=log⁡2xlog⁡28=log⁡2x3\log_8 x=\dfrac{\log_2x}{\log_28}=\dfrac{\log_2x}3; log⁡4x=log⁡2x2\log_4x=\dfrac{\log_2x}2; log⁡2x=log⁡2x\log_2x=\log_2x.

Step 2. Let t=log⁡2xt=\log_2x. The equation becomes t3+t2+t=11\dfrac t3+\dfrac t2+t=11.

Step 3. Common denominator 6: 2t6+3t6+6t6=11t6=11⇒t=6\dfrac{2t}6+\dfrac{3t}6+\dfrac{6t}6=\dfrac{11t}6=11\Rightarrow t=6.

Step 4. So log⁡2x=6⇒x=26=64\log_2x=6\Rightarrow x=2^6=64.

✓Final answer

x=64x=64.

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