Question

Hello everyone, please solve this question.
Prove that !
$\displaystyle\rm 2log \frac{35}{192} + 2log \frac{114}{91} + 2log \frac{13}{19} + log48 = log \frac{75}{64}$

Thanks ! ​

Answer 1

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

We have to prove that:

$\rm \longrightarrow 2 \log \dfrac{35}{192} + 2 \log \dfrac{114}{91} + 2 \log \dfrac{13}{19} + \log48 = \log \dfrac{75}{64}$

Taking LHS, we get:

$\rm = 2 \log \dfrac{35}{192} + 2 \log \dfrac{114}{91} + 2 \log \dfrac{13}{19} + \log48$

$\rm = 2 \bigg[ \log \dfrac{35}{192} + \log \dfrac{114}{91} +\log \dfrac{13}{19} \bigg] + \log48$

We know that:

$\rm \longmapsto \log(x)+ \log(y) +\log(z) + \cdot \cdot \cdot = \log(xyz\cdot \cdot \cdot )$

Therefore, we have:

$\rm = 2 \log \bigg( \dfrac{35}{192} \times \dfrac{114}{91} \times \dfrac{13}{19} \bigg) + \log48$

$\rm = 2 \log \bigg( \dfrac{35}{192} \times \dfrac{114}{7} \times \dfrac{1}{19} \bigg) + \log48$

$\rm = 2 \log \bigg( \dfrac{5}{192} \times \dfrac{114}{1} \times \dfrac{1}{19} \bigg) + \log48$

$\rm = 2 \log \bigg( \dfrac{5}{192} \times 6\bigg) + \log48$

$\rm = 2 \log \bigg( \dfrac{5}{32} \bigg) + \log48$

$\rm = \log \bigg( \dfrac{5}{32} \bigg)^{2} + \log48$

$\rm = \log \bigg( \dfrac{5 \times 5}{32 \times 32} \bigg) + \log48$

$\rm = \log \bigg( \dfrac{5 \times 5}{32 \times 32} \times 48 \bigg)$

$\rm = \log \bigg( \dfrac{5 \times 5}{32 \times 2} \times3\bigg)$

$\rm = \log \bigg( \dfrac{25}{64} \times3\bigg)$

$\rm = \log \bigg( \dfrac{75}{64} \bigg)$

$\rm = RHS$

Hence Proved..!!

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

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

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

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

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

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

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

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

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

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

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