# Thread: Solving for Variable in Exponent.

1. ## 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.

2. $\displaystyle \displaystyle \left|4^{\frac{1}{n}} - 1\right| < x$

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

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

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

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

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

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

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

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

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

4. $\displaystyle ln(1-x)<\frac{2ln2}{n}<ln(1+x)$

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

that is $\displaystyle \frac{1}{ln(1+x)}<\frac{n}{2ln2}<\frac{1}{ln(1-x)}$-- which is nothing but what you want to prove.

5. 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.

6. There are several difficulties with this problem and the replies to it.

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

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

If $\displaystyle x=1$ then the solutions are $\displaystyle n\ge 3.$

7. Post Script
I should have finished this question. (I was put out at those mistakes.)

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

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

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

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

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

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

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

9. 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 $\displaystyle \ln(4)<n\ln(x+1)$ then because $\displaystyle \ln(x+1)>0$ is a fixed positive number we get $\displaystyle n>\dfrac{\ln(4)}{\ln(x+1)}.$

10. Ok, thanks a lot for the help everybody