How do you solve $log x + log 5 = 1$ $logx+log5=1$ ?

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How do you solve $log x + log 5 = 1$ $logx+log5=1$ ?

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Answer 1 · 2016-04-15T01:04:18

$x = 2$ $x = 2$ (Assuming $log$ $Log$ is ${log}_{10}$ $Log_ 10$ )

Explanation:

$log (x) + log (5) = 1$ $Log(x) + Log(5) = 1$

Applying $log (a) + log (b) = log (a b)$ $Log(a) + Log(b) = Log(ab)$
$\to log (5 x) = 1$ $-> Log(5x) = 1$
Assuming $log$ $Log$ is ${log}_{10}$ $Log_10$
$\to 5 x = 10^{1}$ $-> 5x = 10^1$
$5 x = 10$ $5x = 10$
$x = 2$ $x = 2$

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How do you solve $log x + log 5 = 1$ $logx+log5=1$ ?

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