Alberta Free Tutoring And Homework Help For Math 30-1

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  write  $2\log_bx-\frac{\log_bZ}{2}$2 logbx logbZ2   as a single logarithm

6 years ago

Answered By Qi (Jeff) J

Use the identity  $n\log_bx=\log_bx^n$nlogbx=logbxn 

We have:

 $\log_bx^2-\log_bz^{0.5}$logbx2logbz0.5 

Then use the identity:  $\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$logbxlogby=logb(xy ) 

We then have:

 $\log_b\left(\frac{x^2}{z^{0.5}}\right)$logb(x2z0.5 ) 


5 years ago

Answered By Sujalakshmy V

using the identity nlogba = logban

2 log bX=logbX2

logbZ/2 =(1/2)logbZ = logbZ0.5

Hence, 2 log bX - logbZ/2 = logbX-logbZ0.5

Again  using  the identity, logba - logbc = logb(a/c)

logbX-logbZ0.5  = logb(x2/Z0.5)