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

Question

5875 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 : $ln (8x)^(1/2) + ln (4x^2) - ln(16x)^(1/2)$

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

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

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

$=> ln (2sqrt2x^2)$

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

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

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