Question #90b6f

1 Answer
Jul 22, 2016

Answer:

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