Question

What is x if `ln(3x^2) + ln(x^4) + ln(7) =0`?

Answer 1

Use the logarithm laws.

`ln(3x^2 * x^4 * 7) = 0`

`21x^6 = e^0`

`x^6 = 1/21`

`x = +-root(6)(1/21)`

Hopefully this helps!

Answer 2

The solutions are `x=+-root6(1/21)`.

(or `x=+-21^(-1/6)`.)

Explanation:

Use this logarithm rule:

`log_color(green)a(color(red)x)+log_color(green)a(color(blue)y)=log_color(green)a(color(red)x*color(blue)y)`

Here's this rule applied to our equation:

`ln(color(red)(3x^2))+ln(color(blue)(x^4))+ln(color(green)7)=0`

`ln(color(red)(3x^2)*color(blue)(x^4))+ln(color(green)7)=0`

`ln(color(red)3color(purple)(x^6))+ln(color(green)7)=0`

`ln(color(red)3color(purple)(x^6)*color(green)7)=0`

`ln(color(brown)21color(purple)(x^6))=0`

`log_e(color(brown)21color(purple)(x^6))=0`

Convert to exponential form:

`e^0=21x^6`

`1=21x^6`

`1/21=x^6`

`root6(1/21)=x`

Since the root is an even power, we add a plus-or-minus sign:

`x=+-root6(1/21)`

`x=+-root6(21^-1)`

`x=+-(21^-1)^(1/6)`

`x=+-21^(-1/6)`

You can check using a graphing calculator:


Since the values of the zeroes are the same as our answer, we are correct. Hope this helped!