Predicting number of digits with logarithms

[Kartik.]

If log₁₀3 =0.477 then the number of digits in 3⁴⁰ will be?

10^{0.477 = 3
10(0.47740)=340
hm...?}

 

[haruspex]

if log₁₀3 =0.477 then the number of digits in 3⁴⁰ will be?

10^{0.477 = 3
10(0.47740)=340
hm...?}

^{340 = (100.477)40 = 100.477*40}

 

[chiro]

If log₁₀3 =0.477 then the number of digits in 3⁴⁰ will be?

10^{0.477 = 3
10(0.47740)=340
hm...?}

<sup>Hey Kartik and welcome to the forums.

Using log laws, we know that we wish to find log_10(3^40) which gives us the number of decimal digits required to represent 3^40.

Using log laws we first apply log_10(3^40) = 40log_10(3). Now you are given log_10(3), but in future if you wish to find log_b(x) where b is not the natural base e you use the relatioship log_b(x) = ln(x)/ln(b) where ln(x) is the natural logarithm of x and ln(b) must be non-zero (i.e. b can't be 1 or close enough to it for practical purposes).</sup>