How do you approximate log_5 (10/9) given log_5 2=0.4307 and log_5 3=0.6826?

Mar 3, 2018

Approximately $0.0655$

Explanation:

The Logarithmic Division Rule states that:

${\log}_{b} \left(\frac{x}{y}\right) = {\log}_{b} \left(x\right) - {\log}_{b} \left(y\right)$

The Logarithmic Multiplication Rule states that:

${\log}_{b} \left(x \cdot y\right) = {\log}_{b} \left(x\right) + l o {b}_{b} \left(y\right)$

We can apply the logarithmic division rule, so:

${\log}_{5} \left(\frac{10}{9}\right)$

becomes:

${\log}_{5} \left(10\right) - {\log}_{5} \left(9\right)$

We can simplify the numbers in the brackets into the products of prime numbers:

$\to {\log}_{5} \left(2 \cdot 5\right) - {\log}_{5} \left(3 \cdot 3\right)$

Now, we can apply the logarithmic multiplication rule, so:

$\to \left({\log}_{5} \left(2\right) + {\log}_{5} \left(5\right)\right) - \left({\log}_{5} \left(3\right) + {\log}_{5} \left(3\right)\right)$

Now, we can substitute in the values given to us:

$\to \left(0.4307 + 1\right) - \left(0.6826 + 0.6826\right)$

$\left({\log}_{b} \left(b\right) = 1\right)$, always

$= \left(1.4307\right) - \left(1.3652\right)$

$= 0.0655$

Mar 3, 2018

We apply

1. The Logarithmic Division Rule which states that

$\log \left(\frac{x}{y}\right) = \log x - \log y$

2. The Logarithmic Multiplication Rule

$\log \left(x \cdot y\right) = \log x + \log y$

3. And the Logarithmic Power Rule

$\log \left({x}^{y}\right) = y \log x$

Given number can be written as

${\log}_{5} \left(\frac{10}{9}\right)$
$\implies {\log}_{5} \left(\frac{2 \times 5}{3} ^ 2\right)$
$\implies {\log}_{5} \left(2 \times 5\right) - {\log}_{5} {3}^{2}$ .......(Division Rule)
$\implies {\log}_{5} 2 + {\log}_{5} 5 - {\log}_{5} {3}^{2}$ .......(Multiplication Rule)
$\implies {\log}_{5} 2 + {\log}_{5} 5 - 2 {\log}_{5} 3$ .......(Power Rule)

Inserting the given values and remembering that ${\log}_{b} \left(b\right)$ is always$= 1$ we get

${\log}_{5} \left(\frac{10}{9}\right) = 0.4307 + 1 - 2 \times 0.6826$
$\implies {\log}_{5} \left(\frac{10}{9}\right) = 0.0655$