How do you express #log_2 5# in terms of common logs?

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Answer 1 · 2018-04-05T07:16:54

#color(blue)((log_(10)5)/(log_(10)2)#

I am assuming by common logs this means base 10.

If:

#y=log_(b)a<=>b^y=a#

Suppose we wish to express this using a different base. Let's say to a base #bbc#.

From:

#b^y=a#

Take logarithms to the base #c# of both sides:

#ylog_(c)b=log_(c)a#

Divide by #log_(c)b#:

#y=(log_(c)a)/(log_(c)b)#

From above:

#y=log_(b)a#

#:.#

#log_(b)a=(log_(c)a)/(log_(c)b)#

This is the change of base formula:

#log_(2)5=(log_(10)5)/(log_(10)2#