#### Exp

$\left[\left(\frac{B\cdot 3^{\frac{1}{\left(x-1\right)}}}{1+3^{\frac{1}{\left(x-1\right)}}}\right)^x+3\left(\frac{B}{1+3^{\frac{1}{\left(x-1\right)}}}\right)^x\right]^{\frac{1}{x}}$

This is as far as I can get...
$B\left[\left(\frac{3^{\frac{1}{\left(x-1\right)}}}{1+3^{\frac{1}{\left(x-1\right)}}}\right)^x+3\left(\frac{1}{1+3^{\frac{1}{\left(x-1\right)}}}\right)^x\right]^{\frac{1}{x}}$

I can take the B out because it is subject to the same exponent in both parts but I'm at a loss what to do now?

I should be able to get to this expression but I just can't see how - does anyone have any tips for me?

$B\cdot 3^{\frac{1}{x}}\left(1+3^{\frac{1}{\left(x-1\right)}}\right)^{\frac{1}{x}-1}$

Thanks so much!

#### Debsta

MHF Helper
LaTex takes too long for me, sorry.
So here's what I did.
Distributed the x index to numerator and denominator of each term. Leaves 3 on numerator of second fraction
Add fractions - there's a common denominator.
Distribute index of 1/x to numerator and denominator
Here's the tricky bit - factorise out 3^(1/x) from numerator and tidy up indices
See if that works for you

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#### Exp

Thank you so much Debsta

I can get as far as the "tricky" bit now - see below - but I am stuck again!!

$B\left[\left(\frac{\left(3^{\frac{x}{x+1}}+3\right)^{\frac{1}{x}}}{1+3^{\frac{1}{\left(x-1\right)}}}\right)\right]$

Edit: Not sure if the LaTeX is working correctly here but it should be 1/x as the exponent in the numerator.

I must admit I have never encountered questions where I have to take $3^{\frac{1}{x}}$ out of an expression like $\left(3^{\frac{x}{\left(x+1\right)}}+3\right)^{\frac{1}{x}}$ before - is there a method to this because I would love to know!!

#### Debsta

MHF Helper
Ok so far so good. Should be x-1 not x+1 on the first index of 3 though.

So just looking at the top bracket $$\displaystyle 3^{\frac{x}{x-1}} + 3$$
This becomes $$\displaystyle 3* (3^{\frac{x}{x-1} -1} +1)$$ when you factorise out a 3.

All of this is raised to the power of 1/x so now distribute the power and tidy the algebra.
You'll then have the same expression as on the denominator and so can bring them together using basic index law.
Give that a go. Worked for me.

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#### DenisB

Played around with that: ended up needing a couple Tylenols Extra Strength!

In order to eliminate tripping all over the place:
1: cancel out the B's
2: let u = 3^[1 / (x - 1)] and v = 1 + u

Then your equation becomes: [(u/v)^x + 3 * (1/v)^x]^(1/x)
and the given simplified version becomes 3^(1/x) * v^(1/x - 1)

They DO equal each other: try it with assigning a value to x:
I assigned x = 5 : each equation yields .636241955....

The simplified version can be further simplified to: (3v)^(1/x) / v

I'll play with that some more after the hockey game I'm watching!
You do the same...see what you can do...

#### Exp

Hi Debsta and Denis,

At last I got it!!

I removed the 3^1/x expression from the numerator and then distributed the 1/x on the outside bracket to the remaining numerator and denominator and discovering that I had the same expression for both the numerator and denominator but to different powers - I simply subtracted the powers from each other to get the required expression!

Delighted - thank you so much for your support!