How to solve this equation without using In?

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1 Answer
Apr 25, 2018

#a=0.544#

Explanation:

Using the log base rule:
#log_b(c)=log_a(c)/log_a(b)#

#ln()# is just #log_e()#, however, we can use anything else.

#alog_2(7)=3-log_2(14)/log_2(6)#

#alog_2(7)=(3log_2(6)-log_2(14))/log_2(6)#

#alog_2(7)=log_2(6^3/14)/log_2(6)#

#a=log_2(108/7)/(log_2(6)log_2(7))~~0.544#

This has been done without #ln()# however, your spec would probably want you to use #ln()#. Using #ln()# works in a similar way to this, but converting #log_2(7)# to #ln7/ln2# and #log_6(14)# to #ln14/ln6#