How do you express #log_3 10# in terms of common logs?

1 Answer
Dec 30, 2016

Answer:

I changed base to base #10#:

Explanation:

You can change base, remembering that if you want to change base, say, from #b# to #c#, you have:
#log_b(x)=(log_c(x))/(log _c(b))#
In your case we can write:
#log_3(10)=(log_(10)(10))/(log_(10)(3)#
Changing into base #10#.

Now you can evaluate the numerator by yourself....try it!