Question

I need help pls


`log10 \binom{30}{10} - 2 log10 \binom{5}{9} + log10 ( (400)/(343?))`
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Answer 1

`= log_(10) 11.33\\`

Explanation:

Given the expression;

`log_(10)((30)/(10) ) - 2log_(10) (5)/(9) + log_(10)((400)/(343) )`

Using the law of logarithm;

loga + logb = log(ab) and;

log a - log b = log(a/b)

The expression becomes;

`= log_(10)((30)/(10) ) + log_(10)((400)/(343) )- 2log_(10) (5)/(9)\\= log_(10)((30)/(10) ) + log_(10)((400)/(343) )- log_(10) ((5)/(9))^2\\= log_(10)((30)/(10) ) + log_(10)((400)/(343) )- log_(10) (25)/(81)\\= log_(10)((30)/(10) * (400)/(343) / (25)/(81))\\= log_(10)((30)/(10) * (400)/(343) * (81)/(25))\\= log_(10)((3 * 16 * 81)/(343) )\\= log_(10) (3,888)/(343) \\= log_(10) 11.33\\`

Answer 2

9514 1404 393

Answer:

= log(3888/343)

= log(3888) -log(343)

= 4·log(2) +5·log(3) -3·log(7)

≈ 1.054432

Explanation:

Perhaps you want to simplify and evaluate the logarithm.

The applicable rules are ...

log(a/b) = log(a) -log(b)

n·log(a) = log(a^n)

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We will use "log" for "log10". So, your logarithm can be written as ...

log(30/10) -2·log(5/9) +log(400/343)

= log(3) +log(81/25) +log(400/343)

= log(3·81·400/(25·343)) = log(3888/343)

= log(3888) -log(343)

= log(2^4·3^5) -log(7^3) = 4·log(2) +5·log(3) -3·log(7) ≈ 1.054432

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Additional comment

My personal favorite form is the log of a fraction, as it requires the fewest calculator keystrokes. Perhaps the "simplest" is the weighted sum of the logs of primes.