Question

Question #90b6f

Answer 1

Here's my take on this.

Explanation:

You're essentially dealing with a system of two equations with two unknowns, `x` and `y`

`{( (1/2)^(x+y) = 16), (log_(x+y)8 = -3) :}`

Your goal here is to simplify these equations to find another relationship between `x` and `y`.

Start with the first equation

`(1/2)^(x+y) = 16`

Notice that the right side of the equation can be written as

`16 = 2 * 2 * 2 * 2 = 2^4`

Similarly, you know that

`1/2 = 2^(-1)`

This means that the equation is equivalent to

`(2^(-1))^(x+y) = 2^4`

`2^(-x-y) = 2^4 implies -x-y = 4`

This is of course equivalent to

`x + y = -4`

At this point, you should stop because it's clear that you can't find the values of `x` and `y`. Notice that in the second equation, the log has `(x+y)` as the base.

As you know, logarithms must have a positive number that is not equal to `1` as a base. In this case,

`x+y = -4`

does not satisfy this condition, which means that your system of equations has no solution.

You can arrive to the same conclusion, i.e. that you can't find the value of `x` and `y`, by starting with the second equation.

`log_(x+y)8 = -3`

By definition, this will be equivalent to

`(x+y)^(-3) = 8`

Since you know that

`8 = 2 * 2 * 2 = 2^3`

you can write

`(1/(x+y))^(3) = 2^3`

Take the cube root of both sides to find

`root(3)( (1/(x+y))^3) = root(3)(2^3)`

`1/(x+y) = 2`

`2x + 2y = 1 implies x + y = 1/2`

As you can see, this contradicts the value you get for `x+y` by simplifying the first equation, which once again gets you

`color(red)(cancel(color(black)({(x + y = -4), (x + y = 1/2) :}))) ->` the system has no solution