Question

What is x if `3ln2+ln(x^2)+2=4`?

Answer 1

`x=e^{1-3/2 ln(2)}`

Explanation:

Isolate the term involving `x`:

`ln(x^2) = 4-2-3ln(2) = 2-3ln(2)`

Use the property of logarithm `ln(a^b)=bln(a)`:

`2ln(x)=2-3ln(2)`

Isolate the term involving `x` again:

`ln(x)=1-3/2 ln(2)`

Take the exponential of both terms:

`e^{ln(x)} = e^{1-3/2 ln(2)}`

Consider the fact that exponential and logarithm are inverse functions, and thus `e^{ln(x)} =x`

`x=e^{1-3/2 ln(2)}`

Answer 2

`x=+-(esqrt2)/4`

Explanation:

`[1]" "3ln2+ln(x^2)+2=4`

Subtract `2` from both sides.

`[2]" "3ln2+ln(x^2)+2-2=4-2`

`[3]" "3ln2+ln(x^2)=2`

Property: `alog_bm=log_bm^a`

`[4]" "ln2^3+ln(x^2)=2`

`[5]" "ln8+ln(x^2)=2`

Property: `log_bm+log_bn=log_b(mn)`

`[6]" "ln(8x^2)=2`

`[7]" "log_e(8x^2)=2`

Convert to exponential form.

`[8]" "hArre^2=8x^2`

Divide both sides by `8`.

`[9]" "e^2/8=x^2`

Subtract `e^2/8` from both sides.

`[10]" "x^2-e^2/8=0`

Difference of two squares.

`[11]" "(x+sqrt(e^2/8))(x-sqrt(e^2/8))=0`

`[12]" "(x+e/(2sqrt2))(x-e/(2sqrt2))=0`

Rationalize.

`[13]" "(x+(esqrt2)/4)(x-(esqrt2)/4)=0`

Therefore: `color(blue)(x=+-(esqrt2)/4)`