The number of digits in 20301 (given log102 = 0.3010) is

Question

The number of digits in 20301 (given log102 = 0.3010) is (A) 602 (B) 301 (C) 392 (D) 391

Tardigrade Admin · Accepted Answer

Let y=20301 Number of digits = Integral part of (301 log 10 20)+1 = Integral part of [301( log 10 10+ log 10 2)]+1 = Integral part of [301(1+0.3010)]+1 = Integral part of [301 × 1.3010]+1 = Integral part of [391.601]+ 1 =391+1=392

Q. The number of digits in $2 0^{301}$ (given $lo g_{10} 2 = 0.3010$ ) is

602

301

392

391

Let $y = 2 0^{301}$
Number of digits $=$ Integral part of $(301 lo g_{10} 20) + 1$
$=$ Integral part of $[301 (lo g_{10} 10 + lo g_{10} 2)] + 1$
$=$ Integral part of $[301 (1 + 0.3010)] + 1$
$=$ Integral part of $[301 \times 1.3010] + 1$
$=$ Integral part of $[391.601] + 1 = 391 + 1 = 392$

Q. The number of digits in 20301 (given log10​2=0.3010) is

602

301

392

391

Let y=20301 Number of digits = Integral part of (301log10​20)+1 = Integral part of [301(log10​10+log10​2)]+1 = Integral part of [301(1+0.3010)]+1 = Integral part of [301×1.3010]+1 = Integral part of [391.601]+1=391+1=392

Q. The number of digits in $2 0^{301}$ (given $lo g_{10} 2 = 0.3010$ ) is

Let $y = 2 0^{301}$
Number of digits $=$ Integral part of $(301 lo g_{10} 20) + 1$
$=$ Integral part of $[301 (lo g_{10} 10 + lo g_{10} 2)] + 1$
$=$ Integral part of $[301 (1 + 0.3010)] + 1$
$=$ Integral part of $[301 \times 1.3010] + 1$
$=$ Integral part of $[391.601] + 1 = 391 + 1 = 392$