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lnA=lny-lnz+2lnz-3lnx

 Oct 23, 2014

Best Answer 

 #1
avatar+23254 
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A multiplier in front of a log "goes inside" the log as an exponent: 

             Example:  2ln(z)  =  ln(z²)

             Example:  3ln(x)  =  ln(x³)

You subtract ln when your initial problem was division; you add when your initial problem was multiplication:

l(A)  =  ln(y) - ln(z) + 2ln(z) - 3ln(x)

       =  ln(y) - ln(z) + ln(z²) - ln(x³)

       =  ln(y/z) + ln(z²) - ln(x³)

       =  ln( y/z · z² ) - ln(x³)

       =  ln( y · z ) - ln(x³)

       =  ln( y · z / x³ )

 Oct 23, 2014
 #1
avatar+23254 
+5
Best Answer

A multiplier in front of a log "goes inside" the log as an exponent: 

             Example:  2ln(z)  =  ln(z²)

             Example:  3ln(x)  =  ln(x³)

You subtract ln when your initial problem was division; you add when your initial problem was multiplication:

l(A)  =  ln(y) - ln(z) + 2ln(z) - 3ln(x)

       =  ln(y) - ln(z) + ln(z²) - ln(x³)

       =  ln(y/z) + ln(z²) - ln(x³)

       =  ln( y/z · z² ) - ln(x³)

       =  ln( y · z ) - ln(x³)

       =  ln( y · z / x³ )

geno3141 Oct 23, 2014

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