Question

How do you solve `2^ { - 2x + 3} = 8^ { - x }`?

Answer 1

`x=-3`

Explanation:

By using logarithms. Logarithms have many properties that are interesting in solving problems like this. We will focus on three particular ones here:

1) `log_n n=1`
2) `log_n n^y=y`
3) `log_n x^y = y*log_n x`

So, in context:
- property 1 tells us that a log of base `n` of that same number is equal to 1.
- property 2 tells us that a log of base `n` of `n` raised to `y` is equal to `y`.
- property 3 tells us that a log of base `n` of `x` raised to `y` lets us remove that exponent to the front of the log. Now lets apply all that to our equation.

First we "log" both sides, using base 2:

`log_2 2^(-2x+3)=log_2 8^-x`

Using property 3 lets us throw both exponents to the front:

`(-2x+3)*log_2 2=-x*log_2 8`

Now we can rewrite 8 as 2 to the power of 3 and use the first rule on the left side's log and the second rule on the right side's log, leaving us with:

`(-2x+3)*1=-x*3`

Now we just solve for x:

`-2x+3=-3x`
`3x-2x=-3`
`x=-3`