How do you calculate ${log 10}^{3} + log 10 (y^{2})$ $log10^3 + log10(y^2)$ ?

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How do you calculate ${log 10}^{3} + log 10 (y^{2})$ $log10^3 + log10(y^2)$ ?

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Answer 1 · 2016-09-20T18:12:15

$4 + {log}_{10} y^{2}$ $color(green)(4+log_10 y^2)$

Explanation:

Since the default base for $log$ $log$ is $10$ $10$

${log 10}^{3} = {log}_{10} 10^{3} = 3$ $log 10^3=log_10 10^3 = 3$

and remembering that ${log}_{b} p q = {log}_{b} p + {log}_{b} q$ $log_b pq =log_b p +log_b q$
$log 10 (y^{2}) = {log}_{10} 10 + {log}_{10} (y^{2}) = 1 + {log}_{10} (y^{2})$ $log 10(y^2)=log_10 10 +log_10 (y^2)=1 +log_10 (y^2)$

Therefore
${log 10}^{3} + log 10 (y^{2}) = 4 + {log}_{10} (y^{2})$ $log 10^3 + log 10(y^2) = 4 + log_10 (y^2)$

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How do you calculate ${log 10}^{3} + log 10 (y^{2})$ $log10^3 + log10(y^2)$ ?

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