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Solution 3.3:6c

From Förberedande kurs i matematik 1

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Before we even start thinking about transforming log2 and log3 to ln, we use the log laws

logablog(ab)=bloga=loga+logb

to simplify the expression

log3log23118=log3(118log23)=log3118+log3log23.

With help of the relation 2log2x=x and 3log3x=x and taking the natural logarithm , we can express log2 and log3 using ln,

log2x=lnxln2 and log3x=lnxln3.

The two terms log3118 and log3log23 can therefore be written as

log3118=ln3ln118 and log3log23=log3ln2ln3

where we can simplify the last expression further with the logarithm law, log (a/b) = log a – log b, and then transform log3 to ln,

log3ln2ln3=log3ln3log3ln2=ln3lnln3ln3lnln2.

In all, we thus obtain

log3log23118=ln3ln118+ln3lnln3ln3lnln2.

Input into the calculator gives

log3log231184.762.


Note: The button sequence on the calculator will be:


1
  
1
  
8
  
LN
  
÷
  
3
  
LN
  
+
3
  
LN
  
LN
  
÷
  
3
  
LN
  
-
  
2
LN
  
LN
  
÷
  
3
  
LN
  
=