If $log_{10} x = 0.6979$, what is the number of digits in $x^{6979}$ before the decimal point?

Asked Mar 20 '22 at 04:51

Active Mar 20 '22 at 04:51

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Is there a way we can find the number of digits before the decimal point?

My solution approach :-
$log_{10} x = 0.6979$
Let $N=x^{6979}$
$\Rightarrow log_{10}N=6979 \times log_{10} x$
$\Rightarrow log_{10}N=6979 \times 0.6979$
$\Rightarrow log_{10}N=4870.6441$

But I am not aware of any method using which we can find the number of digits before the decimal point in $N$. Please help on this !!!

Thanks in advance !!!

asked Mar 20 '22 at 04:51

Ganit

1,689

2

Note that $N = 10^{\log_{10}N}$, which means $10^{4870} \lt N \lt 10^{4871}$. – John Omielan Mar 20 '22 at 04:58
FYI, there's also Questions on proofs that a number's digits = $\lfloor \log_{10} n \rfloor +1$? (from the bottom of the current right side "Related" section of this question). – John Omielan Mar 20 '22 at 05:05

If $log_{10} x = 0.6979$, what is the number of digits in $x^{6979}$ before the decimal point?

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