Find $x$ in a logarithmic equation

Question

Find $x$ in $\log((x+2)^2) = 2$ where $x > 0$

I began with

$$10^{\log\left((x+2)^2\right)} = 10^2$$

$$(x+2)^2= 100$$

I can do the algebra,but I don't how to apply the restriction.

Thanks.

Answer 1

$$x^2+4x+4=100$$

$$x^2+4x-96=0$$

$$(x-8)(x+12)=0$$

So, $x=8$ and $x=-12$.

$8>0$, but $-12 \ngtr 0$, so applying the restriction, $x=8$

Answer 2

I'm not sure how you multiplied by $10$ on one side, while performed exponentiation on the other in the first step. Did you mean:

$$10^{\log\left((x+2)^2\right)} = 10^2$$ $$(x+2)^2 = 100 \rightarrow x^2+4x-96 =0\rightarrow (x+12)(x-8)=0$$

To get $x=8$ as the solution since $-12<0$. Regardless, you can also apply some logarithm properties to avert the quadratic completely:

$$\log\left((x+2)^2\right)=2\log(x+2)=2$$

Thus giving:

$$\log(x+2)=1$$

And to get rid of the logarithm (assuming base $10$):

$$10^{\log(x+2)}=10^1\rightarrow x+2=10$$

So $x=8$, which is greater than $0$.

Answer 3

$$(x+2)^2 = 100$$

$$x+2 = \pm 10$$ $$x=-2\pm10=-12 \text{ or} 8$$

Since $x>0$, reject $x=-12$ as a solution and hence $x=8$.

Answer 4

I assume this is base $10$ logarithmic we are talking here. If so, $\log_{10}(x+2)^2 = 2$ means $(x+2)^2 = 10^2$. Similarly, $\ln(x+2)^2=\log_{e}(x+2)^2 = 2$ means $(x+2)^2 = e^2$. The rest is just algebra.