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find-the-value-of-x-a-4-2x-1-5-x-2-6-1-x-b-4-3-2x-1-17-3-x-7-0-




Question Number 9753 by j.masanja06@gmail.com last updated on 31/Dec/16
find the value of x  (a)  4^(2x+1) . 5^(x−2) =6^(1−x)   (b)   4(3^(2x+1) )+17(3^x )−7=0
$${find}\:{the}\:{value}\:{of}\:{x} \\ $$$$\left({a}\right)\:\:\mathrm{4}^{\mathrm{2}{x}+\mathrm{1}} .\:\mathrm{5}^{{x}−\mathrm{2}} =\mathrm{6}^{\mathrm{1}−{x}} \\ $$$$\left({b}\right)\:\:\:\mathrm{4}\left(\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{{x}} \right)−\mathrm{7}=\mathrm{0} \\ $$
Commented by ridwan balatif last updated on 31/Dec/16
(b) 4(3^(2x+1) )+17(3^x )−7=0          4(3^(2x) ×3)+17(3^x )−7=0              12(3^x )^2 +17(3^x )−7=0         let 3^x =a, then      12a^2 +17a−7=0  (3a−1)(4a+7)=0  a=(1/3)                    a=((−7)/4)  3^x =3^(−1)               3^x =((−7)/4)(IMPOSSIBLE)  x=−1
$$\left(\mathrm{b}\right)\:\mathrm{4}\left(\mathrm{3}^{\mathrm{2x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{\mathrm{x}} \right)−\mathrm{7}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}\left(\mathrm{3}^{\mathrm{2x}} ×\mathrm{3}\right)+\mathrm{17}\left(\mathrm{3}^{\mathrm{x}} \right)−\mathrm{7}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{12}\left(\mathrm{3}^{\mathrm{x}} \right)^{\mathrm{2}} +\mathrm{17}\left(\mathrm{3}^{\mathrm{x}} \right)−\mathrm{7}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{let}\:\mathrm{3}^{\mathrm{x}} =\mathrm{a},\:\mathrm{then} \\ $$$$\:\:\:\:\mathrm{12a}^{\mathrm{2}} +\mathrm{17a}−\mathrm{7}=\mathrm{0} \\ $$$$\left(\mathrm{3a}−\mathrm{1}\right)\left(\mathrm{4a}+\mathrm{7}\right)=\mathrm{0} \\ $$$$\mathrm{a}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}=\frac{−\mathrm{7}}{\mathrm{4}} \\ $$$$\mathrm{3}^{\mathrm{x}} =\mathrm{3}^{−\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{x}} =\frac{−\mathrm{7}}{\mathrm{4}}\left(\mathrm{IMPOSSIBLE}\right) \\ $$$$\mathrm{x}=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by ridwan balatif last updated on 31/Dec/16
(a)  4^(2x+1) .5^(x−2) =6^(1−x)   4^(2x) ×4×5^x ×(1/(25))=(6/6^x )        16^x ×(2/(75))×5^x =(1/6^x )                        480^x =((75)/2)            log(480^x )=log(((75)/2))          x.log480    =log(((75)/2))                                 x=log_(480) (((75)/2))                                 x≈0.587
$$\left(\mathrm{a}\right)\:\:\mathrm{4}^{\mathrm{2x}+\mathrm{1}} .\mathrm{5}^{\mathrm{x}−\mathrm{2}} =\mathrm{6}^{\mathrm{1}−\mathrm{x}} \\ $$$$\mathrm{4}^{\mathrm{2x}} ×\mathrm{4}×\mathrm{5}^{\mathrm{x}} ×\frac{\mathrm{1}}{\mathrm{25}}=\frac{\mathrm{6}}{\mathrm{6}^{\mathrm{x}} } \\ $$$$\:\:\:\:\:\:\mathrm{16}^{\mathrm{x}} ×\frac{\mathrm{2}}{\mathrm{75}}×\mathrm{5}^{\mathrm{x}} =\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{x}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{480}^{\mathrm{x}} =\frac{\mathrm{75}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{log}\left(\mathrm{480}^{\mathrm{x}} \right)=\mathrm{log}\left(\frac{\mathrm{75}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{x}.\mathrm{log480}\:\:\:\:=\mathrm{log}\left(\frac{\mathrm{75}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{log}_{\mathrm{480}} \left(\frac{\mathrm{75}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\approx\mathrm{0}.\mathrm{587} \\ $$

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