How do you evaluate ${log}_{6} x + {log}_{6} 3 = 2$ $log_(6)x + log_(6)3 = 2$ ?

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How do you evaluate ${log}_{6} x + {log}_{6} 3 = 2$ $log_(6)x + log_(6)3 = 2$ ?

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Answer 1 · 2016-04-21T19:16:28

$x = 12$ $x=12$

Explanation:

Use Property: ${log}_{b} (x y) = {log}_{b} x + {log}_{b} y and y = {log}_{b} x \Leftrightarrow b^{y} = x$ $log _b(xy)=log_bx+log_by and y=log_bx iff b^y=x$

${log}_{6} (3 x) = 2$ $log_6(3x)=2$

$6^{2} = 3 x$ $6^2 = 3x$

$\frac{36}{3} = x$ $36/3=x$

$x = 12$ $x=12$

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How do you evaluate ${log}_{6} x + {log}_{6} 3 = 2$ $log_(6)x + log_(6)3 = 2$ ?

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How do you evaluate ${log}_{6} x + {log}_{6} 3 = 2$ $log_(6)x + log_(6)3 = 2$ ?