# Question #90b6f

Jul 22, 2016

Here's my take on this.

#### Explanation:

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

$\left\{\begin{matrix}{\left(\frac{1}{2}\right)}^{x + y} = 16 \\ {\log}_{x + y} 8 = - 3\end{matrix}\right.$

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

${\left(\frac{1}{2}\right)}^{x + y} = 16$

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

$16 = 2 \cdot 2 \cdot 2 \cdot 2 = {2}^{4}$

Similarly, you know that

$\frac{1}{2} = {2}^{- 1}$

This means that the equation is equivalent to

${\left({2}^{- 1}\right)}^{x + y} = {2}^{4}$

${2}^{- x - y} = {2}^{4} \implies - x - y = 4$

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 $\left(x + y\right)$ 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

${\left(x + y\right)}^{- 3} = 8$

Since you know that

$8 = 2 \cdot 2 \cdot 2 = {2}^{3}$

you can write

${\left(\frac{1}{x + y}\right)}^{3} = {2}^{3}$

Take the cube root of both sides to find

$\sqrt[3]{{\left(\frac{1}{x + y}\right)}^{3}} = \sqrt[3]{{2}^{3}}$

$\frac{1}{x + y} = 2$

$2 x + 2 y = 1 \implies x + y = \frac{1}{2}$

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

$\textcolor{red}{\cancel{\textcolor{b l a c k}{\left\{\begin{matrix}x + y = - 4 \\ x + y = \frac{1}{2}\end{matrix}\right.}}} \to$ the system has no solution