Question

How do you solve `8^(x-2) = 32^(x+10)`?

Answer 1

`x` = -28

Explanation:

The first step to solving this equation is changing the bases so that they are equal. To do this, find a number that can be raised to an exponent to equal each base. In this case, `2` can be raised to the third power to equal `8` and raised to the fifth power to equal `32`:

`2^3` = 8 | `2^5` = 32

Leaving us with:

`(2^3)^(x-2)` = `(2^5)^(x+10)`

Using a special property of exponents, we can multiply the exponents on each side of the equation:

3(`x` - 2) = 3`x` - 6 | 5(`x` + 10) = 5`x` + 50

Leaving us with:

`2^(3x-6)` = `2^(5x+50)`

Now that the bases are equal, we know that the exponents are also equal:

3`x` - 6 = 5`x` + 50

Now all we need to do is solve for `x` using algebra:

2`x` = -56
`x` = -28**