How do you solve $log 2 + log x = 1$ $log2+logx=1$ ?

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Answer 1 · 2016-07-12T01:22:55

5 (Assuming $log = {log}_{10}$ $log = log_10$ )

${log}_{10} 2 + {log}_{10} x = 1$ $log_10 2 + log_10 x = 1$

Using: ${log}_{n} a + {log}_{n} b = {log}_{n} (a b)$ $log_n a + log_n b = log_n(ab)$ we may rewrite the equation as:

${log}_{10} 2 x = 1$ $log_10 2x = 1$

Using: ${log}_{n} a = b \to a = n^{b}$ $log_n a = b -> a = n^b$ we may simplify the equation as:

$2 x = 10^{1}$ $2x = 10^1$

$2 x = 10$ $2x = 10$

$x = 5$ $x = 5$

How do you solve log2+logx=1log2+logx=1log2+logx=1?