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Question 68 of 69

Q.Find all real numbers satisfying the equation : 4x−3(2x+2)+25=04^x-3(2^{x+2})+2^5=0.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★
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Substitutes t=2xt=2^x to turn the exponential equation into a quadratic, solves it, then converts back to xx.

  1. Rewrite each term in base 22: 4x=(22)x=(2x)24^x=(2^2)^x=(2^x)^2, and 2x+2=2x⋅22=4⋅2x2^{x+2}=2^x\cdot2^2=4\cdot2^x, and 25=322^5=32.
  2. Let t=2xt=2^x (note t>0t>0 for all real xx). The equation becomes t2−3(4t)+32=0⇒t2−12t+32=0t^2-3(4t)+32=0\Rightarrow t^2-12t+32=0.
  3. Solve using the quadratic formula: t=12±144−1282=12±162=12±42t=\dfrac{12\pm\sqrt{144-128}}2=\dfrac{12\pm\sqrt{16}}2=\dfrac{12\pm4}2, giving t=8t=8 or t=4t=4. …

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