Question

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

Answer 1

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

Explanation:

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

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

Explanation:

`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`