How do you simplify #ln(8x)^(1/2)+ln4x^2-ln(16x)^(1/2)#?

Question

5595 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-17T14:45:09

#ln(8x)^(1/2)+ln4x^2-ln(16x)^(1/2)=2lnx+3/2ln2#

We will use identities #lna+lnb=ln(a*b)#, #lna-lnc=ln(a/c)# and #lna^n=nlna#.

Hence #ln(8x)^(1/2)+ln4x^2-ln(16x)^(1/2)#

= #ln(((8x)^(1/2)*4x^2)/(16x)^(1/2))#

= #ln((8^(1/2)color(red)(x^(1/2))*4x^2)/(16^(1/2)color(red)(x^(1/2))))#

= #ln((8^(1/2)*4x^2)/16^(1/2))#

= #ln((2^2x^2)/2^(1/2))#

= #ln2^2+lnx^2-ln2^(1/2)#

= #2ln2+2lnx-1/2ln2#

= #2lnx+3/2ln2#

Answer 2 · 2018-05-17T14:59:57

#color(indigo)(=> (3/2) ln 2 + 2 ln x#

#log m + log n = log (mn)#

#log m - log n = log (m/n)#

Given : #ln (8x)^(1/2) + ln (4x^2) - ln(16x)^(1/2)#

#=> ln ((8x)^(1/2) * (4x^2)) - ln (16x)^(1/2)#

#=> ln ((2sqrt2*4*x^(1/2)*x^2)/(16x)^(1/2))#

#=> ln ((cancel(8)^2sqrt2 * cancel((x)^(1/2)) * x^2) / (cancel4 * cancel((x)^(1/2))))#

#=> ln (2sqrt2x^2)#

#=> ln (2^(3/2)) + ln x^2#

#color(indigo)(=> (3/2) ln 2 + 2 ln x#