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How do you rewrite #log_(1/5)x# as a ratio of common logs and natural logs?

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Feb 23, 2017

#log_b a=log_c a/log_c b#
#log_(1/5) x=log_10 x/log_10 (1/5)#
#log_(1/5) x=ln x/ln (1/5)#

Explanation:

You would apply the formula:

#log_b a=log_c a/log_c b#

Then, for example,

#log_(1/5) x=log_10 x/log_10 (1/5)#

and

#log_(1/5) x=ln x/ln (1/5)#

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