What is x if $3 ln 2 + ln (x^{2}) + 2 = 4$ $3ln2+ln(x^2)+2=4$ ?

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What is x if $3 ln 2 + ln (x^{2}) + 2 = 4$ $3ln2+ln(x^2)+2=4$ ?

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Answer 1 · 2015-11-08T14:06:37

$x = e^{1 - \frac{3}{2} ln (2)}$ $x=e^{1-3/2 ln(2)}$

Explanation:

Isolate the term involving $x$ $x$ :

$ln (x^{2}) = 4 - 2 - 3 ln (2) = 2 - 3 ln (2)$ $ln(x^2) = 4-2-3ln(2) = 2-3ln(2)$

Use the property of logarithm $ln (a^{b}) = b ln (a)$ $ln(a^b)=bln(a)$ :

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

Isolate the term involving $x$ $x$ again:

$ln (x) = 1 - \frac{3}{2} ln (2)$ $ln(x)=1-3/2 ln(2)$

Take the exponential of both terms:

$e^{ln (x)} = e^{1 - \frac{3}{2} ln (2)}$ $e^{ln(x)} = e^{1-3/2 ln(2)}$

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

$x = e^{1 - \frac{3}{2} ln (2)}$ $x=e^{1-3/2 ln(2)}$

Answer 2 · 2015-11-08T14:07:20

$x = \pm \frac{e \sqrt{2}}{4}$ $x=+-(esqrt2)/4$

Explanation:

$[1] 3 ln 2 + ln (x^{2}) + 2 = 4$ $[1]" "3ln2+ln(x^2)+2=4$

Subtract $2$ $2$ from both sides.

$[2] 3 ln 2 + ln (x^{2}) + 2 - 2 = 4 - 2$ $[2]" "3ln2+ln(x^2)+2-2=4-2$

$[3] 3 ln 2 + ln (x^{2}) = 2$ $[3]" "3ln2+ln(x^2)=2$

Property: $a {log}_{b} m = {log}_{b} m^{a}$ $alog_bm=log_bm^a$

$[4] {ln 2}^{3} + ln (x^{2}) = 2$ $[4]" "ln2^3+ln(x^2)=2$

$[5] ln 8 + ln (x^{2}) = 2$ $[5]" "ln8+ln(x^2)=2$

Property: ${log}_{b} m + {log}_{b} n = {log}_{b} (m n)$ $log_bm+log_bn=log_b(mn)$

$[6] ln (8 x^{2}) = 2$ $[6]" "ln(8x^2)=2$

$[7] {log}_{e} (8 x^{2}) = 2$ $[7]" "log_e(8x^2)=2$

Convert to exponential form.

$[8] \Leftrightarrow e^{2} = 8 x^{2}$ $[8]" "hArre^2=8x^2$

Divide both sides by $8$ $8$ .

$[9] \frac{e^{2}}{8} = x^{2}$ $[9]" "e^2/8=x^2$

Subtract $\frac{e^{2}}{8}$ $e^2/8$ from both sides.

$[10] x^{2} - \frac{e^{2}}{8} = 0$ $[10]" "x^2-e^2/8=0$

Difference of two squares.

$[11] (x + \sqrt{\frac{e^{2}}{8}}) (x - \sqrt{\frac{e^{2}}{8}}) = 0$ $[11]" "(x+sqrt(e^2/8))(x-sqrt(e^2/8))=0$

$[12] (x + \frac{e}{2 \sqrt{2}}) (x - \frac{e}{2 \sqrt{2}}) = 0$ $[12]" "(x+e/(2sqrt2))(x-e/(2sqrt2))=0$

Rationalize.

$[13] (x + \frac{e \sqrt{2}}{4}) (x - \frac{e \sqrt{2}}{4}) = 0$ $[13]" "(x+(esqrt2)/4)(x-(esqrt2)/4)=0$

Therefore: $x = \pm \frac{e \sqrt{2}}{4}$ $color(blue)(x=+-(esqrt2)/4)$

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