How do you evaluate #Ln 432 = y#?

1 Answer
Apr 3, 2017

Answer:

#ln 432 ~~ 6.0684#

Explanation:

Suppose you know:

#log 2 ~~ 0.30103#

#log 3 ~~ 0.47712#

#ln 10 ~~ 2.302585#

Then:

#(ln 432)/(ln 10) = log 432 = log (2^4 3^3) = 4 log 2 + 3 log 3#

So:

#ln 432 = (ln 10)(4 log 2 + 3 log 3)#

#color(white)(ln 432) ~~ 2.302585*(4*0.30103 + 3*0.47712)#

#color(white)(ln 432) ~~ 6.0684#