How do you solve $ln (5 - 2 x^{2}) + ln 9 = ln 43$ $ln(5-2x^2)+ln9=ln43$ ?

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How do you solve $ln (5 - 2 x^{2}) + ln 9 = ln 43$ $ln(5-2x^2)+ln9=ln43$ ?

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Answer 1 · 2018-05-27T16:59:24

$x = \pm \frac{1}{3}$ $x=pm1/3$

Explanation:

We have
$ln (5 - 2 x^{2}) = ln (43) - ln (9)$ $ln(5-2x^2)=ln(43)-ln(9)$

$ln (5 - 2 x^{2}) = ln (\frac{43}{9})$ $ln(5-2x^2)=ln(43/9)$

so

$5 - 2 x^{2} = \frac{43}{9}$ $5-2x^2=43/9$

$5 - \frac{43}{9} = 2 x^{2}$ $5-43/9=2x^2$

$\frac{2}{9} = 2 x^{2}$ $2/9=2x^2$

$\frac{1}{9} = x^{2}$ $1/9=x^2$

Answer 2 · 2018-05-27T17:05:51

See explanation below

Explanation:

The goal is to get a expresion $log A = log B$ $logA=logB$ . By inyectivity of logarithm we arrive to $A = B$ $A=B$

In our case: $ln (5 - 2 x^{2}) + ln 9 = ln 43$ $ln(5-2x^2)+ln9=ln43$

$ln (5 - 2 x^{2}) = ln 43 - ln 9 = \frac{ln 43}{9}$ $ln(5-2x^2)=ln43-ln9=ln43/9$

Then $5 - 2 x^{2} = \frac{43}{9}$ $5-2x^2=43/9$ or equivalent

$5 - \frac{43}{9} = 2 x^{2}$ $5-43/9=2x^2$

$\frac{2}{9} = 2 x^{2}$ $2/9=2x^2$

$\frac{1}{9} = x^{2}$ $1/9=x^2$

$x = \pm \frac{1}{3}$ $x=+-1/3$

It is obvious that both solution are valid

Answer 3 · 2018-05-27T17:08:21

$x = \pm \frac{1}{3}$ $x = +- 1/3$

Explanation:

Given: $ln (5 - 2 x^{2}) + ln 9 = ln 43$ $ln (5 - 2x^2) + ln 9 = ln 43$

Use the logarithm property $ln a + ln b = ln (a \cdot b)$ $ln a + ln b = ln (a*b)$

$ln ((5 - 2 x^{2}) 9) = ln 43$ $ln ((5 - 2x^2)9) = ln 43$

$ln (45 - 18 x^{2}) = ln 43$ $ln (45 - 18x^2) = ln 43$

Exponentiate both sides and use the logarithm property $e^{ln x} = x$ $e^ln x = x$

$e^{ln (45 - 18 x^{2})} = e^{ln 43}$ $e^(ln (45 - 18x^2)) = e^(ln 43)$

$45 - 18 x^{2} = 43$ $45 - 18x^2 = 43$

$45 - 43 = 18 x^{2}$ $45 - 43 = 18x^2$

$2 = 18 x^{2}$ $2 = 18 x^2$

$\frac{2}{18} = \frac{1}{9} = x^{2}$ $2/18 = 1/9 = x^2$

$x = \pm \frac{1}{3}$ $x = +- 1/3$

CHECK to see if the answers work in the problem (must be ln of a positive number)#:

$ln (5 - 2 {(\frac{1}{3})}^{2}) = ln (5 - \frac{2}{9}) > 0$ $ln (5-2(1/3)^2) = ln (5-2/9) > 0$

$ln (5 - 2 {(- \frac{1}{3})}^{2}) = ln (5 - \frac{2}{9}) > 0$ $ln (5-2(-1/3)^2) = ln (5-2/9) > 0$

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How do you solve $ln (5 - 2 x^{2}) + ln 9 = ln 43$ $ln(5-2x^2)+ln9=ln43$ ?

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