Question

Question #072fe

Answer 1

Please see the explanation.

Explanation:

I am going to assume base 10:

`log_10(root(3)(10(x-1)^2)) - 1/3log_10((x+3)^2) = 1/3`

Remove the root from inside the first logarithm by multiplying by 1/3:

`1/3log_10(10(x-1)^2) - 1/3log_10((x+3)^2) = 1/3`

Multiply both sides of the equation by 3:

`log_10(10(x-1)^2) - log_10((x+3)^2) = 1`

The difference of two logarithms is the same as division within the argument:

`log_10(10(x-1)^2/(x+3)^2) = 1`

Eliminate the logarithm by making both sides the power of 10:

`10^(log_10(10(x-1)^2/(x+3)^2)) = 10^1`

On the left side the base and the logarithm are removed:

`10(x-1)^2/(x+3)^2 = 10^1`

Divide both sides by 10:

`(x-1)^2/(x+3)^2 = 1`

Multiply both sides by `(x+3)^2`:

`(x-1)^2=(x+3)^2`

Expand the squares:

`x^2 - 2x + 1 = x^2 + 6x + 9`

Add `-x^2 -6x - 1` to both sides:

`-8x = 8`

Divide by sides by -8:

`x = -1`

Answer 2

`x=-5/4`

Explanation:

`logroot3(10(x-1)^2)-1/3log(3+x)^2=1/3`

`logroot3(10(x-1)^2)-log(3+x)^(2/3)`

`log(root3(10(x-1)^2)/(3+x)^(2/3))=1/3`

`root3(10(x-1)^2)/(3+x)^(2/3)=10^(1/3)`

`root3(10(x-1)^2)=10^(1/3)((3+x)^(2/3))`

`(root3(10(x-1)^2))^3=(10^(1/3)((3+x)^(2/3)))^3`

`10(x-1)^2=10(3+x)^2`

`(x-1)^2=(3+x)^2`

`x^2-2x-1=x^2+6x+9`

`8x+10=0`

`8x=-10`

`x=-10/8=-5/4`