Question

Log 10^800 = 2 + 3. Log 10^ 2​

Answer 1

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

We have to prove the given equality.

Taking LHS, we get:

$\rm = log_{10}(800)$

$\rm = log_{10}(8 \times 100)$

$\rm = log_{10}(100) + log_{10}(8)$

$\rm = log_{10}( {10}^{2} ) + log_{10}( {2}^{3} )$

$\rm = 2 \log_{10}(10) + 3 \log_{10}(2 )$

$\rm = 2 \times 1 + 3 \log_{10}(2 )$

$\rm = 2 + 3 \log_{10}(2 )$

$\rm = RHS$

Hence Proved..!!

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

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

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

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

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

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

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

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

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

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

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