Question

pls solve the problem. It's from fundamental of maths​

Answer 1

Step-by-step explanation:

We have,

$(log_{10}(x) ) ^{2} + log_{10}(x)^{2} - \{( log_{10}(2) )^{2} - 1 \} = 0$

$\implies \: (log_{10}(x) ) ^{2} + 2 log_{10}(x) - \{( log_{10}(2) )^{2} - 1 \} = 0$

Let $log_{10}(x) = y$

Then,

$\: y ^{2} + 2y - \{( log_{10}(2) )^{2} - 1 \} = 0$

$\implies\: y = \frac{ - 2 \pm \sqrt{4+ 4 {( log_{10}(2) )}^{2} - 4 } }{2}$

$\implies\: y = \frac{ - 2 \pm \sqrt{4 {( log_{10}(2) )}^{2}} }{2}$

$\implies\: y = \frac{ - 2 \pm 2 log_{10}(2) }{2}$

$\implies\: y = - 1\pm log_{10}(2)$

$\implies\: y = - 1 + log_{10}(2) \: \: or \: \: y = - 1 - log_{10}(2)$

$\implies\: x = 10 ^{- 1 + log_{10}(2) } \: \: or \: \: x =10 ^{ - 1 - log_{10}(2) }$

$\implies\: x = 10 ^{- 1} .10^{ log_{10}(2)} \: \: or \: \: x =10 ^{ - 1 }.10 ^{ - log_{10}(2) }$

$\implies\: x = 2 \times 10 ^{- 1} \: \: or \: \: x =10 ^{ - 1} \times {2}^{ - 1}$

$\implies\: x = \frac{1}{5} \: \: or \: \: x = \frac{1}{20}$