How close?

Sep 2009
181
0
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 2010
354
173
\(\displaystyle (x+3)^4\) is non-negative, so we can just move some stuff around:


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

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

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

\(\displaystyle \implies~ \frac{1}{10} > |x+3|\)
 
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Sep 2009
181
0
\(\displaystyle (x+3)^4\) is non-negative, so we can just move some stuff around:


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

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

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

\(\displaystyle \implies~ \frac{1}{10} > |x+3|\)
Great thanks!

I just have a couple of rookie questions.

If it was negative, you couldn't move stuff around?

and why did the (x+3) become an absolute value?
 
Jan 2010
354
173
Great thanks!

I just have a couple of rookie questions.

If it was negative, you couldn't move stuff around?

and why did the (x+3) become an absolute value?
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.
 
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