Question

If log2=o.3010 and log3 =0.4771,find value of log5-log27+log under root 8

Answer 1

log5-log27+log√8

= log 10/2 - log (3)^3 + log2✓2

= log10- log2- 3log3+log2✓2

=1-0.3010-3(0.4771)+(0.1651)

=1-0.3010-1.4313-0.1651

=-0.5612

I hope you understand...

Answer 2

Therefore the value of log 5 - log 27 + log √8 is -0.2806.

Given:

Logarithmic values: log 2 = 0.3010, log 3 = 0.4771, and log 5 = 0.6989

To Find:

The value of log 5 - log 27 + log √8

Solution:

The given question can be solved as shown below.

Given that: log 2 = 0.3010, log 3 = 0.4771, and log 5 = 0.6989

Let y = log 5 - log 27 + log √8

⇒ y = log 5 - log 3³ + log 2√2

⇒ y = log 5 - 3log 3 + log 2√2      [ ∵ log aⁿ = n log a ]

⇒ y = log 5 - 3log 3 + log 2 + log √2      [ ∵ log ab = log a + log b ]

⇒ y = log 5 - 3log 3 + log 2 + log $2^1^/^2$

⇒ y = log 5 - 3log 3 + log 2 + 1/2 log 2       [ ∵ log aⁿ = n log a ]

Now substituting the values of log 2, log 3, and log 5 in the above equation.

⇒ y = log 5 - 3log 3 + log 2 + 1/2 log 2

⇒ y = 0.6989 - ( 3 × 0.4771 ) + 0.3010 + ( 1/2 × 0.3010 )

⇒ y = 0.6989 - 1.4313 + 0.3010 + 0.1505

⇒ y = 1.1507 - 1.4313

⇒ y = -0.2806

Hence y = log 5 - log 27 + log √8 = -0.2806

Therefore the value of log 5 - log 27 + log √8 is -0.2806.

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