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Exercise 3.5 · Q6

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

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Step 1. Rewrite everything in base 22. 4x=(22)x=(2x)24^x=(2^2)^x=(2^x)^2; 2x+2=4⋅2x2^{x+2}=4\cdot2^x; 25=322^5=32. The equation becomes (2x)2−3(4⋅2x)+32=0(2^x)^2-3(4\cdot2^x)+32=0, i.e. (2x)2−12(2x)+32=0(2^x)^2-12(2^x)+32=0.

Step 2. Substitute y=2xy=2^x. y2−12y+32=0y^2-12y+32=0. …

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