3^(4-x) =< 3^x

Let us use logarithm:

Since log is an increasing function, then :

log 3 ^(4-x) =< log 3^x

We know that: log a^b = blog a

==> (4-x) log 3 =< x log 3

Divide by log 3 ( log 3 > 0 so ot does not change the direction of the inequality)

==> 4-x =< x

Now add x to both sides:

==> 4 =< 2x

Divide by 2:

==> 2 =< x

Then the solution is :

x belong to the interval [2, inf)

To solve 3^(4-x) = <3^x

Solution:

Given 3^(4-x) = < 3^x. Both sides are the exponents of the same base and so the relation holds for the exponets also by one to one property.

So 4-x = < x

Add x to both sides:

4 = < x+x

4 = < 2x

Divide by 2:

2 = < x

So x >= 2.

This is an exponential inequality.

3^(4-x) =< 3^x

Since the bases are the same and they are >1, the direction of the inequality still holds and we'll use the one to one property:

4-x =< x

We'll subtract x both sides:

4-x-x=< x-x

4-2x=<0

We'll subtract 4 both sides:

-2x =< -4

We'll divide by -2 both sides. After the division, the direction of the inequality will be changed.

x >= -4/-2

x >= 2

**The interval of admissible values for x is [2, +infinite).**