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Question Number 202203 by SEKRET last updated on 22/Dec/23

                       2^(2023) = abc.............^(______________)              a+b+c = ?

$$\:\: \\ $$$$ \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2023}} =\:\overset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_} {\boldsymbol{\mathrm{abc}}.............} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\:=\:? \\ $$$$ \\ $$$$ \\ $$$$\: \\ $$$$ \\ $$

Answered by Frix last updated on 22/Dec/23

With a calculator not capable of exact  numbers as great as ≈10^(609) :  log 2^(2023)  =2023log 2  log 2 ≈.301029995664  2023log 2 ≈608.983681228  10^(.983681228) ≈9.63121833  2^(2023) =963...  9+6+3=18

$$\mathrm{With}\:\mathrm{a}\:\mathrm{calculator}\:\mathrm{not}\:\mathrm{capable}\:\mathrm{of}\:\mathrm{exact} \\ $$$$\mathrm{numbers}\:\mathrm{as}\:\mathrm{great}\:\mathrm{as}\:\approx\mathrm{10}^{\mathrm{609}} : \\ $$$$\mathrm{log}\:\mathrm{2}^{\mathrm{2023}} \:=\mathrm{2023log}\:\mathrm{2} \\ $$$$\mathrm{log}\:\mathrm{2}\:\approx.\mathrm{301029995664} \\ $$$$\mathrm{2023log}\:\mathrm{2}\:\approx\mathrm{608}.\mathrm{983681228} \\ $$$$\mathrm{10}^{.\mathrm{983681228}} \approx\mathrm{9}.\mathrm{63121833} \\ $$$$\mathrm{2}^{\mathrm{2023}} =\mathrm{963}... \\ $$$$\mathrm{9}+\mathrm{6}+\mathrm{3}=\mathrm{18} \\ $$

Commented by SEKRET last updated on 23/Dec/23

   Hello, thank you sir for your reply. Can it be done without a calculator? Is there an easier way?

$$\: \\ $$Hello, thank you sir for your reply. Can it be done without a calculator? Is there an easier way?

Commented by Frix last updated on 23/Dec/23

I don′t see any other way.

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{any}\:\mathrm{other}\:\mathrm{way}. \\ $$

Commented by SEKRET last updated on 23/Dec/23

thank you sir Frix

$$\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{Frix}} \\ $$

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