# Solving for Variable in Exponent.

• March 21st 2011, 06:27 PM
Relmiw
Solving for Variable in Exponent.
I'm trying to solve for n in the following inequality

|4^(1/n) - 1| < x

where n is a natural number and x is a positive real number. I feel like i should be able to do this but i can't think of how. Could someone please help me out with this?

I'm also at a loss for how to solve for n with

100^n / n! < x

but I think i might be able to figure that out once i know how to work with the first one.
• March 21st 2011, 06:31 PM
Prove It
$\displaystyle \left|4^{\frac{1}{n}} - 1\right| < x$

$\displaystyle -x < 4^{\frac{1}{n}} - 1 < x$

$\displaystyle 1 - x < 4^{\frac{1}{n}} < 1 + x$

$\displaystyle \ln{(1 - x)} < \ln{\left(4^{\frac{1}{n}}\right)} < \ln{(1 + x)}$

$\displaystyle \ln{(1 - x)} < \frac{1}{n}\ln{4} < \ln{(1 + x)}$

$\displaystyle \ln{(1 - x)} < \frac{2\ln{2}}{n} < \ln{(1 + x)}$

$\displaystyle \frac{1}{\ln{(1 + x)}} < \frac{n}{2\ln{2}} < \frac{1}{\ln{(1 - x)}}$

$\displaystyle \frac{2\ln{2}}{\ln{(1 + x)}} < n < \frac{2\ln{2}}{\ln{(1 - x)}}$.
• March 22nd 2011, 07:33 AM
Relmiw
Quote:

Originally Posted by Prove It
$\displaystyle \ln{(1 - x)} < \frac{2\ln{2}}{n} < \ln{(1 + x)}$

$\displaystyle \frac{1}{\ln{(1 + x)}} < \frac{n}{2\ln{2}} < \frac{1}{\ln{(1 - x)}}$

How is it that you got from this first step to the second? I've tried everything i can think of but i think there is some log rule I don't know about.
• March 22nd 2011, 08:29 AM
Sambit
$ln(1-x)<\frac{2ln2}{n}

$\frac{1}{ln(1-x)}>\frac{n}{2ln2}>\frac{1}{ln(1+x)}$ (taking reciprocal)

that is $\frac{1}{ln(1+x)}<\frac{n}{2ln2}<\frac{1}{ln(1-x)}$-- which is nothing but what you want to prove.
• March 22nd 2011, 10:44 AM
HallsofIvy
If 0< a< b< c, then a< b so 1< b/a and 1/b< 1/a; b< c so, by the same divisions 1/c< 1/b. Together those give
1/c< 1/b< 1/a.

If the three numbers are not all positive, it is not true. For example, -2< 1< 2, 1/2< 1 but those are not less than -1/2.
• March 22nd 2011, 11:28 AM
Plato
There are several difficulties with this problem and the replies to it.

If $x>1$ then $\ln(1- x)$ does not exist.
That is one difficulty the proposed solution.

On the other hand, if $0 then $\ln(1-x)<0$ because $0<1-x<1$.
That means that $\ln(1-x)<0, another difficulty as pointed out in reply #5.

If $x=1$ then the solutions are $n\ge 3.$
• March 22nd 2011, 02:33 PM
Plato
Post Script
I should have finished this question. (I was put out at those mistakes.)

First note that for all $n$ we have $4^{\frac{1}{n}}>1$.
If $x\ge 1$ then $1-x<0<4^{\frac{1}{n}}.
So this is true if $n\ge 3$.

In the case $0 we have $0<1-x<1<4^{\frac{1}{n}} <1+x$.

Now we can use the $\ln$ function.
We get $\ln(4) or $n>\dfrac{\ln(4)}{\ln(x+1)}$.
• March 22nd 2011, 03:44 PM
Relmiw
Quote:

Originally Posted by Plato
Post Script
I should have finished this question. (I was put out at those mistakes.)

First note that for all $n$ we have $4^{\frac{1}{n}}>1$.
If $x\ge 1$ then $1-x<0<4^{\frac{1}{n}}.
So this is true if $n\ge 3$.

In the case $0 we have $0<1-x<1<4^{\frac{1}{n}} <1+x$.

Now we can use the $\ln$ function.
We get $\ln(4) or $n>\dfrac{\ln(4)}{\ln(x+1)}$.

Thanks for the help so far everyone. I've been trying to wrap my head around them for a while now

Regarding when 0 < x < 1, i see how we get to 1 - x < 1 < 4^(1/n) < 1 + x
but when we use ln, don't we get n > ln(4) / ln(x + 1) > (n * ln(1 - x)) / (ln(1 + x) ?
when you used ln you dropped the last part for some reason.
• March 22nd 2011, 03:59 PM
Plato
Quote:

Originally Posted by Relmiw
Regarding when 0 < x < 1, i see how we get to 1 - x < 1 < 4^(1/n) < 1 + x but when we use ln, don't we get n > ln(4) / ln(x + 1) > (n * ln(1 - x)) / (ln(1 + x) ?when you used ln you dropped the last part for some reason.

The 'last part' makes no different whatsoever.
If $\ln(4) then because $\ln(x+1)>0$ is a fixed positive number we get $n>\dfrac{\ln(4)}{\ln(x+1)}.$
• March 22nd 2011, 07:01 PM
Relmiw
Ok, thanks a lot for the help everybody