How do you write as a single log for ${log}_{6} 60 - {log}_{6} 10$ $log_6 60 - log_6 10$ ?

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How do you write as a single log for ${log}_{6} 60 - {log}_{6} 10$ $log_6 60 - log_6 10$ ?

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Answer 1 · 2018-03-16T10:45:19

${log}_{6} 6 = 1$ $log_6 6=1$

Explanation:

$using the laws of logarithms$ $"using the "color(blue)"laws of logarithms"$

$∙ x log x - log y = log (\frac{x}{y})$ $•color(white)(x)logx-logy=log(x/y)$

$∙ x {log}_{b} b = 1$ $•color(white)(x)log_b b=1$

$\Rightarrow {log}_{6} 60 - {log}_{6} 10 = {log}_{6} (\frac{60}{10}) = {log}_{6} 6 = 1$ $rArrlog_6 60-log_6 10=log_6(60/10)=log_6 6=1$

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How do you write as a single log for ${log}_{6} 60 - {log}_{6} 10$ $log_6 60 - log_6 10$ ?

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How do you write as a single log for ${log}_{6} 60 - {log}_{6} 10$ $log_6 60 - log_6 10$ ?