Question #070b2

2 Answers
Nov 23, 2016

Assuming that the logarithm is base 10:
#log_10(9) = ln(9)/ln(10)#

Aug 18, 2017

#log_10(9) = (ln(9))/(ln(10))#

Explanation:

Consider #log_10(9)#

Set #" " log_(10)(9) = x#

Another way of writing this is:#" "10^x=9#

Set #" "e^b=9=>color(red)(b=ln(9))#

But we have:#" " e^b=9=10^x#

Take natural logs of both sides and using the principle that
#ln(10^x)# is the same as #xln(10)#

#bln(e)=xln(10)#

but #ln(e)=1#

#b=xln(10)#

Thus #x=(color(red)(b))/(ln(10))#

but #b=ln(9)# giving

#x=ln(9)/(ln(10))#

But we set # log_(10)(9) = x# giving:

#log_10(9) = (ln(9))/(ln(10))#