Question

I need help pls

Answer 1

Answer:

$= log_{10} 11.33$

Step-by-step explanation:

Given the expression;

$log_{10}(\frac{30}{10} ) - 2log_{10} \frac{5}{9} + log_{10}(\frac{400}{343} )$

Using the law of logarithm;

loga + logb = log(ab) and;

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

The expression becomes;

$= log_{10}(\frac{30}{10} ) + log_{10}(\frac{400}{343} )- 2log_{10} \frac{5}{9}\\= log_{10}(\frac{30}{10} ) + log_{10}(\frac{400}{343} )- log_{10} (\frac{5}{9})^2\\= log_{10}(\frac{30}{10} ) + log_{10}(\frac{400}{343} )- log_{10} \frac{25}{81}\\= log_{10}(\frac{30}{10} \times \frac{400}{343} \div \frac{25}{81})\\= log_{10}(\frac{30}{10} \times \frac{400}{343} \times \frac{81}{25})\\= log_{10}(\frac{3 \times 16 \times 81}{343} )\\= log_{10} \frac{3,888}{343} \\= log_{10} 11.33$

Answer 2

9514 1404 393

Answer:

  = log(3888/343)

  = log(3888) -log(343)

  = 4·log(2) +5·log(3) -3·log(7)

  ≈ 1.054432

Step-by-step 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.