Question

Question #9aa2f

Answer 1

`"Part A":` `x = 0`

`"Part B":` `x in RR`

Explanation:

`"Part A":` We have: `(2^(8))^(x) = 1`

The "power rule" of exponents states that `(a^(m))^(n) = a^(m n)`.

What this rule states is that if you raise a power to another power, you must multiply the two exponents together.

So in this case, the `8` and the `x` should be multiplied together to get "`8 x`":

`Rightarrow 2^(8 x) = 1`

Another important rule of exponents is that any number raised to the power `0` is equal to `1`.

Even `2^(0)` is equal to `1`:

`Rightarrow 2^(8 x) = 2^(0)`

Now, there is one more rule we need to follow to solve this equation.

When both sides of the equation have the same base, the exponents are equal to each other.

For example, `a^(2 x) = a^(4)`

We can then say that `2 x = 4`

Let's apply this rule to our case:

`Rightarrow 8 x = 0`

`therefore x = 0`

`"`

`"Part B":` We have: `(5^(0))^(x) = 1`

For this question, we can use the same rules as the previous one.

First, let's multiply the two powers together to get one exponent:

`Rightarrow 5^(0 timesx) = 1`

`Rightarrow 5^(0) = 1`

Then, let's rewrite `1` as `5^(0)`.

`Rightarrow 5^(0) = 5^(0)`

Now, both sides of the equation are exactly the same.

So this equation is true for all real values of `x`.

In other words, `x in RR`.

`"`

Source:

http://www.mesacc.edu/~scotz47781/mat120/notes/exponents/review/review.html