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64^(x-4)=4^2x can someone please explain to me how to solve this
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\[\large 64= 4^3\]\[\Large ((4)^3)^{(x-4)}= 4^{2x}\]\[\Large 4^{3(x-4)}=4^{2x}\] Therefore 3(x-4) = 2x can you do it now?
yeah but that first part confuses me can you tell me like what you did
ok 64 = 4^3 So inorder to make the bases same i replaced 64 with 4^3 Now we know that\[\Large (x^m)^n=x^{m \times n}\]Same way\[\Large (4^3)^{(x-4)} =4^{3(x-4)}\] Is this clear ?
yeah thank you!
Anytime :)
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