# Natural Logarithmic Equation `ln 3 +ln (3x^2) = 4`

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`ln3+ln(3x^2)=4`

To solve, express the left side with one logarithm. To do so, apply the product rule which is `lnm+lnn=ln(m*n)` .

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

`ln(9x^2)=4`

Then, express it in exponential equation. The equivalent exponential equation of `ln m =a` is `m=e^a` .

`9x^2=e^4`

From here, isolate x. To do so, divide both sides by 9.

`(9x^2/)9=e^4/9`

`x^2=e^4/9`

And take the square root of both sides.

`sqrt(x^2)=+-sqrt(e^4/9)`

`x=+-e^2/3`

**Hence, the solution to the given equation is `x={-e^2/3, e^2/3}` .**

given `ln3+ln(3x^2)=4`

Because we are working with natural logarithm, it means always the base is e.

Also we know that `lnm+ln n=ln(mn)`

So, `ln(3.3x^2)=4.1` As `lne=1`

So, `ln (9x^2)=4lne`

or, `ln(9x^2)=lne^4` As `plnm=lnm^p`

or, `9x^2=e^4`

or, `(9x^2-e^4)=0`

or, `((3x)^2-(e^2)^2)=0`

or, `(3x-e^2)(3x+e^2)=0`

Now, `(3x-e^2)=0rArrx=e^2/3` ,

and `(3x+e^2)=0rArrx=-e^2/3` .

We know

`ln(m)+ln(n)=ln(mn)`

`ln(m)=x` then `e^x=m`

Thus we have

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

`ln(3xx3x^2)=4`

`e^4=3xx3x^2`

`9x^2=e^4`

`x^2=e^4/9`

`x=+-sqrt(e^4/9)`

`x=+-e^2/3`