Question 381634
My guess is that the expression is:
{{{4^((log(3, (3^(log(4, (x+1)))))))}}}
(In case you can't read 4's exponent above, it is: {{{log(3, (3^(log(4, (x+1)))))}}}
If this is right, then the simple way to simplify this expression requires that you understand what logarithms represent.<br>
The idea behind logarithms is that <i>any</i> positive number raised to the correct power will result in <i>any</i> other positive number. In general, {{{log(a, (b))}}} represents the "correct exponent" for a that results in b. <br>
Some of these "correct exponents" are easy to find. For example:
{{{log(q, (q)) = 1}}} since 1 is the power for q that results in q.
{{{log(z, (1)) = 0}}} since 0 is the power for z that results in 1.
{{{log(2, (16)) = 4}}} since 4 is the power for 2 that results in 16.
{{{log(25, (5)) = 1/2}}} since 1/2 (i.e. square root) is the power for 25 that results in 5.
{{{log(v, (v^w)) = w}}} since w is the power for v that results in {{{v^x}}}.
(Think about this last one because we will be using it on your expression shortly.)<br>
When you can't figure out the "correct exponent" we can find a decimal approximation for the exponent using our calculators and the base conversion formula (if the logarithm is not base 10 or base e).<br>
Now back to your expression...
The exponent for 4 is:
{{{log(3, (3^(log(4, (x+1)))))}}}
This fits the pattern for {{{log(v, (v^w)) = w}}}
So {{{log(3, (3^(log(4, (x+1))))) = log(4, (x+1))}}} because {{{log(4, (x+1))}}} is the exponent for 3 that results in {{{3^log(4, (x+1))}}} and we find it as an exponent of 3!! So the simplified exponent for 4 is: {{{log(4, (x+1))}}}
Now the full expression is:
{{{4^log(4, (x+1))}}}
And once again we have the pattern for {{{log(v, (v^w)) = w}}}. So 
{{{4^log(4, (x+1)) = x+1}}} since {{{log(4, (x+1))}}} is the exponent for 4 that results in x+1 and we find it as an exponent of 4. Replacing {{{4^log(4, (x+1))}}} with x+1 we get:
x+1