Hello,

Can you please help me figure this out?

Problem: How close to -3 do we have to take x so that

$\displaystyle 1/(x+3)^4 > 10,000$

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- Jan 21st 2010, 11:57 PMl flipboi lHow close?
Hello,

Can you please help me figure this out?

Problem: How close to -3 do we have to take x so that

$\displaystyle 1/(x+3)^4 > 10,000$ - Jan 22nd 2010, 12:06 AMdrumist
$\displaystyle (x+3)^4$ is non-negative, so we can just move some stuff around:

$\displaystyle \frac{1}{(x+3)^4} > 10,000$

$\displaystyle \implies~ \frac{1}{10,000} > (x+3)^4$

$\displaystyle \implies~ \sqrt[4]{\left(\frac{1}{10,000}\right)} > |x+3|$

$\displaystyle \implies~ \frac{1}{10} > |x+3|$ - Jan 22nd 2010, 12:24 AMl flipboi l
- Jan 22nd 2010, 12:34 AMdrumist
Say for example we considered something like $\displaystyle \frac{1}{(x+3)^3} > 1000$ . It's possible for $\displaystyle (x+3)^3$ to be a negative number, but if it was negative, the value on the left-hand side would have been negative, and a negative number can't be greater than 1000. So we just need to make a side note that we are only considering values of $\displaystyle x$ such that $\displaystyle x>-3$. From then we can continue the problem.

As for your second question: Any time you take an even-powered root of a value, you must take its absolute value. In other words:

$\displaystyle a^2 = 25 ~\implies~ \sqrt{a^2}=\sqrt{25} ~\implies~ |a|=5 ~\implies~ a=\pm5$

Notice that had we not done so, we would not have found the solution $\displaystyle a=-5$, but clearly $\displaystyle (-5)^2=25$.

You do not need to do this when taking odd-powered roots.