Menu Close

3-4-5-x-4-6-x-1-dx-1-2-2x-1-find-the-value-of-x-Solution-4-5-x-3-4-6-x-2-2-x-k

Tinku Tara 0 Comments
Pin




Question Number 202636 by ibroclex_adex last updated on 30/Dec/23
((^3 (√4^(5−x) ))/(∫_4 ^6 (x−1)dx)) = (1/2^(2x−1) ) , find the value of x. Solution (4^((5−x)/3) /(∫_4 ^6 ((x^2 /2)−x+k))) = (1/2^(2x−1) ) (2^(2•((5−x)/3)) /(((6^2 /2)−6+k)−((4^2 /2)−4+k))) = (1/2^(2x−1) ) (2^((10−2x)/3) /(((36)/2)−6+k−((16)/2)+4−k)) = (1/2^(2x−1) ) (2^((10−2x)/3) /(18−6−8+4)) = (1/2^(2x−1) ) (2^((10−2x)/3) /8) = (1/2^(2x−1) ) (Cross Multiply) 2^(2x−1) ×2^((10−2x)/3) = 8×1 2^(2x−1) ×2^((10−2x)/3) = 2^3 2^(2x−1+((10−2x)/3)) = 2^3 (Since, the bases are equal. Then, we can equate the exponents) 2x−1+((10−2x)/3) = 3 (Multiply each term by 3) 3(2x)−3(1)+3(((10−2x)/3)) = 3(3) 6x−3+10−2x = 9 (Collect Like Terms) 4x+7 = 9 4x = 9−7 4x = 2 (Divide Both Sides by 4) ((4x)/4) = (2/4) ∴ x = (1/2)
$$\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\:^{\mathrm{3}} \sqrt{\mathrm{4}^{\mathrm{5}−\mathrm{x}} }}{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right){dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\mathrm{Solution}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}^{\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{x}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\mathrm{2}\bullet\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\left(\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{6}+\mathrm{k}\right)−\left(\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\frac{\mathrm{36}}{\mathrm{2}}−\mathrm{6}+\cancel{\mathrm{k}}−\frac{\mathrm{16}}{\mathrm{2}}+\mathrm{4}−\cancel{\mathrm{k}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{18}−\mathrm{6}−\mathrm{8}+\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:\:\left(\mathrm{Cross}\:\mathrm{Multiply}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{8}×\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \:\left(\mathrm{Since},\:\mathrm{the}\:\mathrm{bases}\:\mathrm{are}\:\mathrm{equal}.\:\mathrm{Then},\:\mathrm{we}\:\mathrm{can}\:\mathrm{equate}\:\mathrm{the}\:\mathrm{exponents}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}\:=\:\mathrm{3}\:\left(\mathrm{Multiply}\:\mathrm{each}\:\mathrm{term}\:\mathrm{by}\:\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\mathrm{2x}\right)−\mathrm{3}\left(\mathrm{1}\right)+\cancel{\mathrm{3}}\left(\frac{\mathrm{10}−\mathrm{2x}}{\cancel{\mathrm{3}}}\right)\:=\:\mathrm{3}\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6x}−\mathrm{3}+\mathrm{10}−\mathrm{2x}\:=\:\mathrm{9}\:\left(\mathrm{Collect}\:\mathrm{Like}\:\mathrm{Terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}+\mathrm{7}\:=\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{9}−\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{2}\:\left(\mathrm{Divide}\:\mathrm{Both}\:\mathrm{Sides}\:\mathrm{by}\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\cancel{\mathrm{4}x}}{\cancel{\mathrm{4}}}\:=\:\frac{\mathrm{2}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *