Question

How do you solve `ln(5-2x^2)+ln9=ln43`?

Answer 1

`x=pm1/3`

Explanation:

We have
`ln(5-2x^2)=ln(43)-ln(9)`

`ln(5-2x^2)=ln(43/9)`

so

`5-2x^2=43/9`

`5-43/9=2x^2`

`2/9=2x^2`

`1/9=x^2`

Answer 2

See explanation below

Explanation:

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

In our case: `ln(5-2x^2)+ln9=ln43`

`ln(5-2x^2)=ln43-ln9=ln43/9`

Then `5-2x^2=43/9` or equivalent

`5-43/9=2x^2`

`2/9=2x^2`

`1/9=x^2`

`x=+-1/3`

It is obvious that both solution are valid

Answer 3

`x = +- 1/3`

Explanation:

Given: `ln (5 - 2x^2) + ln 9 = ln 43`

Use the logarithm property `ln a + ln b = ln (a*b)`

`ln ((5 - 2x^2)9) = ln 43`

`ln (45 - 18x^2) = ln 43`

Exponentiate both sides and use the logarithm property `e^ln x = x`

`e^(ln (45 - 18x^2)) = e^(ln 43)`

`45 - 18x^2 = 43`

`45 - 43 = 18x^2`

`2 = 18 x^2`

`2/18 = 1/9 = x^2`

`x = +- 1/3`

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

`ln (5-2(1/3)^2) = ln (5-2/9) > 0`

`ln (5-2(-1/3)^2) = ln (5-2/9) > 0`