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If-a-9999999998000000001-Find-A-a-1-1-9-A-




Question Number 202044 by hardmath last updated on 19/Dec/23
If  a = 9999999998000000001  Find  A = (((√a) + 1))^(1/9)    →   A = ?
$$\mathrm{If} \\ $$$$\mathrm{a}\:=\:\mathrm{9999999998000000001} \\ $$$$\mathrm{Find} \\ $$$$\mathrm{A}\:=\:\sqrt[{\mathrm{9}}]{\sqrt{\mathrm{a}}\:+\:\mathrm{1}}\:\:\:\rightarrow\:\:\:\mathrm{A}\:=\:? \\ $$
Commented by mr W last updated on 19/Dec/23
do you have a calculator? you′ll get  A≈11.364  please check your question again,  maybe you have written one 9 too   much!
$${do}\:{you}\:{have}\:{a}\:{calculator}?\:{you}'{ll}\:{get} \\ $$$${A}\approx\mathrm{11}.\mathrm{364} \\ $$$${please}\:{check}\:{your}\:{question}\:{again}, \\ $$$${maybe}\:{you}\:{have}\:{written}\:{one}\:\mathrm{9}\:{too}\: \\ $$$${much}! \\ $$
Answered by esmaeil last updated on 19/Dec/23
(((√(99999.99999∗99999.99999∗10^8 ))+1))^(1/9) =  ((99999.99999∗10^4 +1))^(1/9) =  ((999999999.9+1))^(1/9) ≅10
$$\sqrt[{\mathrm{9}}]{\sqrt{\mathrm{99999}.\mathrm{99999}\ast\mathrm{99999}.\mathrm{99999}\ast\mathrm{10}^{\mathrm{8}} }+\mathrm{1}}= \\ $$$$\sqrt[{\mathrm{9}}]{\mathrm{99999}.\mathrm{99999}\ast\mathrm{10}^{\mathrm{4}} +\mathrm{1}}= \\ $$$$\sqrt[{\mathrm{9}}]{\mathrm{999999999}.\mathrm{9}+\mathrm{1}}\cong\mathrm{10} \\ $$
Answered by mr W last updated on 20/Dec/23
99^2 =9801  999^2 =998001  ...  999...9^2 _(n times) =(10^n −1)^2 =(10^n −2)×10^n +1=999...9_(n−1) 8000...0_(n−1) 1    a=99999999_(8 times) 800000000_(8 times) 1=999 999 999^2   (√a)=999 999 999  (√a)+1=999 999 999+1=1 000 000 000=10^9   A=(((√a)+1))^(1/9) =((10^9 ))^(1/9) =10 ✓
$$\mathrm{99}^{\mathrm{2}} =\mathrm{9801} \\ $$$$\mathrm{999}^{\mathrm{2}} =\mathrm{998001} \\ $$$$… \\ $$$$\underset{{n}\:{times}} {\mathrm{999}…\mathrm{9}^{\mathrm{2}} }=\left(\mathrm{10}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{10}^{{n}} −\mathrm{2}\right)×\mathrm{10}^{{n}} +\mathrm{1}=\underset{{n}−\mathrm{1}} {\mathrm{999}…\mathrm{9}8}\underset{{n}−\mathrm{1}} {\mathrm{000}…\mathrm{0}1} \\ $$$$ \\ $$$${a}=\underset{\mathrm{8}\:{times}} {\mathrm{99999999}8}\underset{\mathrm{8}\:{times}} {\mathrm{00000000}1}=\mathrm{999}\:\mathrm{999}\:\mathrm{999}^{\mathrm{2}} \\ $$$$\sqrt{{a}}=\mathrm{999}\:\mathrm{999}\:\mathrm{999} \\ $$$$\sqrt{{a}}+\mathrm{1}=\mathrm{999}\:\mathrm{999}\:\mathrm{999}+\mathrm{1}=\mathrm{1}\:\mathrm{000}\:\mathrm{000}\:\mathrm{000}=\mathrm{10}^{\mathrm{9}} \\ $$$${A}=\sqrt[{\mathrm{9}}]{\sqrt{{a}}+\mathrm{1}}=\sqrt[{\mathrm{9}}]{\mathrm{10}^{\mathrm{9}} }=\mathrm{10}\:\checkmark \\ $$
Commented by hardmath last updated on 20/Dec/23
cool dear professor thankyou
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thankyou} \\ $$

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