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Question-56824




Question Number 56824 by Gulay last updated on 24/Mar/19
Commented by Gulay last updated on 24/Mar/19
sir could you help me
$$\mathrm{sir}\:\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19
4^x ×4^2 +3×4^x =76  4^x =a  16a+3a=76  19a=76  a=4   so 4^x =4^1   x=1
$$\mathrm{4}^{{x}} ×\mathrm{4}^{\mathrm{2}} +\mathrm{3}×\mathrm{4}^{{x}} =\mathrm{76} \\ $$$$\mathrm{4}^{{x}} ={a} \\ $$$$\mathrm{16}{a}+\mathrm{3}{a}=\mathrm{76} \\ $$$$\mathrm{19}{a}=\mathrm{76} \\ $$$${a}=\mathrm{4}\:\:\:{so}\:\mathrm{4}^{{x}} =\mathrm{4}^{\mathrm{1}} \\ $$$${x}=\mathrm{1} \\ $$
Answered by kelly33 last updated on 25/Mar/19
4^(x+2) +3.4^x =76  4^x .4^2 +3.4^x =76  Suposition 4^x =y   16y+3y=76  19y=76  y=((76)/(19))  y=4    4^x =4^1   x=1
$$\mathrm{4}^{{x}+\mathrm{2}} +\mathrm{3}.\mathrm{4}^{{x}} =\mathrm{76} \\ $$$$\mathrm{4}^{{x}} .\mathrm{4}^{\mathrm{2}} +\mathrm{3}.\mathrm{4}^{{x}} =\mathrm{76} \\ $$$${Suposition}\:\mathrm{4}^{{x}} ={y}\: \\ $$$$\mathrm{16}{y}+\mathrm{3}{y}=\mathrm{76} \\ $$$$\mathrm{19}{y}=\mathrm{76} \\ $$$${y}=\frac{\mathrm{76}}{\mathrm{19}} \\ $$$${y}=\mathrm{4} \\ $$$$ \\ $$$$\mathrm{4}^{{x}} =\mathrm{4}^{\mathrm{1}} \\ $$$${x}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by Gulay last updated on 25/Mar/19
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$$$ \\ $$

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