Question

$2 log_{10}(5) + log_{10}(8) - \frac{1}{2 log_{10}(4)$
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Answer 1

$2 log_{10}(5) + log_{10}(8) - \frac{1}{2} log_{10}(4)$

$= log_{10}(5^{2}) + log_{10}(8) - \frac{1}{2} log_{10}(2^{2})$

$= log_{10}(25)+ log_{10}(8) - \frac{1}{2} \times 2 log_{10}(2)$

$\boxed {\pink{ n log_{a} x = log_{a} x^{n} }}$

$= log_{10}(25\times 8)- log_{10}(2)$

$\boxed {\pink{ log_{a} x+ log_{a} y = log_{a} (xy) }}$

$= log_{10}\frac{(25\times 8)}{2}$

$\boxed {\pink{ log_{a} x- log_{a} y = log_{a} \Big(\frac{x}{y}\Big) }}$

$= log_{10}(25\times 4)$

$= log_{10}(100)$

$= log_{10}(10^{2})$

$= 2log_{10}(10)$

$= 2 \times 1$

$\boxed {\pink{ log_{a} a = 1 }}$

$= 2$

Therefore.,

$\red{2 log_{10}(5) + log_{10}(8) - \frac{1}{2} log_{10}(4)} \green {= 2}$

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