How do you calculate ${log}_{6} (2)$ $log_6 (2)$ ?

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Answer 1 · 2016-04-29T16:52:38

log 2/log 6 = 0.386853, nearly

Use ${log}_{b} a = \frac{{log}_{c} a}{{log}_{c} b}$ $log_b a = log_c a/log_c b$ , where c is at your choice.

Let us choose c as either 10 or e. Then

${log}_{6} 2 = \frac{{log}_{10} 2}{{log}_{10} 6} = \frac{ln 2}{ln 6} = 0.386853$ $log_6 2=log_10 2/log_10 6=ln 2/ln 6=0.386853$ , nearly.

I am sure that this comparison, using different bases, would reinforce the formula used, in the long term memory.

How do you calculate log6(2)log_6 (2)?