Question

If log2 = 0.30103 and log 3 = 0.4771, find the number of digits in (648)5

Answer 1

Answer:

15 Digits

Step-by-step explanation:

If log2 = 0.30103 and log 3 = 0.4771,

find the number of digits in (648)5

first find the log (648)⁵

log (648)⁵

using log aⁿ = n log a

= 5 log (648)

= 5 log ( 8 * 81)

= 5 log ( 2³ * 3⁴)

log(ab) = log a + log b

= 5 *  ( log ( 2³)  + log ( 3⁴) )

= 5 *( 3 log 2  + 4 log 3)

= 5 * ( 3 * 0.30103  + 4 * 0.4771)

= 5 * (0.90309 + 1.9084)

= 5 * (2.81149)

= 14.05745

Taking

log (648)⁵ = 14.05745

=> (648)⁵ = 10^(14.05745)

10ⁿ has n + 1 digits

here n = 14.05  < 15

=> n + 1 = 15  digits

15 Digits

Answer 2

Answer:

Step-by-step explanation:

Concept:

Power rule of logarithm:

The logarithm of an exponential number is the exponent times the logarithm of the base.

$log (a)^b = b log a$

Product rule of logarithm:

The logarithm of the product is the sum of the logarithms of the factors.

$log (ab) = log a +log b$

Given:

log2 = 0.30103

log 3 = 0.4771

To Find:

Number of digits in  $648^{5}$

Solution:

$x = 648^{5}$

Taking Logarithm on both sides,

$log x = log (648)^5$

using power rule of logarithm,

$log (a)^b = bloga$

$log x = 5 log 648$

$log x = 5 log ( 8 * 81 )$

$log x = 5 log ( 2^3* 3^4 )$

using product rule of logarithm,

$log (ab) = log a +log b$

$log x = 5 log 2^3 + 5log 3^4$

$log x = 15 log 2 + 20log 3$

$log x = 15 * 0.30103 + 20 * 0.4771$

$log x = 4.51545 + 9.542$

$log x = 14.05745$

$log 648^{5} = 14.05745$

$648^5 = 10^{14.05745}$

Number of digits in $10^{n}$ is $n +1$

Since the logarithm of $648^5$ has $14$ as the integral part, the number of digits in the number $648^5$ is $14+1 = 15$.

Hence, the number of digits in $648^5$ is 15

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