If logb2 = x and logb3 = y, evaluate the following in terms of x and y:

You need these three rules:

1.     logb(ac) = logba + logbc

2.     logb = logba - logbc

3.     logb = c·logba


(a) logb216=

Break 216 down to primes:

216 =  =  =  =  =  = 

logb216 = logb 

= logb + logb     (by rule 1.)

= 3·logb2 + 3·logb3                 (by rule 3.)

= 3·x + 3·y                         (by substitution.) 

= 3(x + y) 

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(b) logb162=

Break 162 down to primes:

162 =  =  =  = 

logb162 = logb 

= logb2 + logb     (by rule 1.)

= logb2 + 4·logb3           (by rule 3.)

= x + 4·y                   (by substitution.) 

= x + 4y 

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(c) logb

Break the 9 in the denominator down into primes

9 =  = 

logb = logb

= logb2 - logb

= logb2 - 2·logb3

= x - 2y

============================================

(d) 

 logb9
———————    
 logb8

Break the 9 down as  and the 8 down as 

 logb9     logb
——————— = —————————    
 logb8     logb


 2·logb3
—————————                (by rule 3.)    
 3·logb2

            (substitution)

Edwin