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.
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Do you mean $\log((x+2)^2)$ or $\log^2(x+2)$? – Andrew Li Mar 04 '18 at 04:08
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The first one, let me edit the question. – Alexis Arias Pineda Mar 04 '18 at 04:11
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1Just work out the problem without the restriction, finding all solutions, then apply the restriction at the end. – Parcly Taxel Mar 04 '18 at 04:12
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Why do you multiply the LHS with $10$? Did you mean to write it as $10^{\log((x + 2)^2)}$ instead? – an4s Mar 04 '18 at 04:14
4 Answers
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$$x^2+4x+4=100$$
$$x^2+4x-96=0$$
$$(x-8)(x+12)=0$$
So, $x=8$ and $x=-12$.
$8>0$, but $-12 \ngtr 0$, so applying the restriction, $x=8$
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$$(x+2)^2 = 100$$
$$x+2 = \pm 10$$ $$x=-2\pm10=-12 \text{ or} 8$$
Since $x>0$, reject $x=-12$ as a solution and hence $x=8$.
Siong Thye Goh
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Are you sure we will reject the $x=-12$? In the original problem statement we have $\log((x+2)^2)$ so if $x=-12$ we still get $\log(100)$. I believe if the original problem was written as $2\log(x+2)$ then we would discard the $-12$ – Teh Rod Mar 04 '18 at 05:31
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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 $0$.
Andrew Li
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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.
