write $2\log_bx-\frac{\log_bZ}{2}$2 logbx− logbZ2 as a single logarithm
7 years ago
Use the identity $n\log_bx=\log_bx^n$nlogbx=logbxn
We have:
$\log_bx^2-\log_bz^{0.5}$logbx2−logbz0.5
Then use the identity: $\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$logbx−logby=logb(xy )
We then have:
$\log_b\left(\frac{x^2}{z^{0.5}}\right)$logb(x2z0.5 )
using the identity nlogba = logban
2 log bX=logbX2
logbZ/2 =(1/2)logbZ = logbZ0.5
Hence, 2 log bX - logbZ/2 = logbX2 -logbZ0.5
Again using the identity, logba - logbc = logb(a/c)
logbX2 -logbZ0.5 = logb(x2/Z0.5)
7 years ago
Answered By Qi (Jeff) J
Use the identity $n\log_bx=\log_bx^n$nlogbx=logbxn
We have:
$\log_bx^2-\log_bz^{0.5}$logbx2−logbz0.5
Then use the identity: $\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$logbx−logby=logb(xy )
We then have:
$\log_b\left(\frac{x^2}{z^{0.5}}\right)$logb(x2z0.5 )
7 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 = logbX2 -logbZ0.5
Again using the identity, logba - logbc = logb(a/c)
logbX2 -logbZ0.5 = logb(x2/Z0.5)