Question

Please find value of
log√32​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given logarithm.

$\rm =\log( \sqrt{32})$

$\rm =\log( \sqrt{ {2}^{5} })$

$\rm =\log({2}^{5})^{ \frac{1}{2} }$

$\rm =\log(2)^{ \frac{5 \times 1}{2} }$

Now, we know that:

$\rm = \log_{x}( {a}^{m} ) = m\log_{x}(a)$

Therefore, we get:

$\rm = \dfrac{5}{2} \log(2)$

We know that log(2) ≈ 0.3. Therefore:

$\rm = \dfrac{5}{2} \times 0.3$

$\rm = \dfrac{1.5}{2}$

$\rm =0.75$

Therefore:

$\rm \longrightarrow\log( \sqrt{32}) = 0.75$

★ Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

$\rm 1. \: \: {a}^{n} = b \implies log_{a}(b) = n$

$\rm 2. \: \: log_{a}(1) = 0, \: a \neq0,1$

$\rm 3. \: \: log_{a}(a) = 1, \: a \neq0,1$

$\rm 4. \: \: log_{a}(x) = log_{a}(y) \implies x = y$

$\rm 5. \: \: log_{e}(x) = ln(x)$

$\rm6. \: \: log_{a}(x) + log_{a}(y) = log_{a}(xy)$

$\rm7. \: \: log_{a}(x) - log_{a}(y) = log_{a} \bigg( \dfrac{x}{y} \bigg)$

$\rm 8. \: \: log_{a}( {x}^{n} ) = n\log_{a}(x)$

$\rm 9. \: \: log_{a}(m) = \dfrac{ log_{b}(m) }{ log_{b}(a) },m > 0,b > 0,a \ne1,b \ne1$

$\rm 10. \: \: log_{a}(b) = \dfrac{1}{ log_{b}(a) }$