2300.11 – Exponential Equation With Threes


An equation finds itself drifting among familiar algebraic scenes. It looks around, recalling the expressions of its past— how vague they are now, how long it has spent assuming different forms in order to mean something subtly different: worn its three on the other side, for example, or complicated its operations. With each new manipulation, the memory of former selves pales. What it represents is almost out of reach---like an asymptote. Where does it fit in this cacophany of numbers and symbols? Can it all be nothing more than a series of abstractions?

In other words, help this equation:

32x+23x+33x+3=0,3^{2x+2} - 3^{x+3} - 3^x + 3 = 0,

find its roots (i.e., solve for xx). Please!


Solution

Let's try grouping:

32x+23x+33x+3=0,3x+23x3x+23(3x3)=0,(3x+21)(3x3)=0. \begin{aligned} 3^{2x+2} - 3^{x+3} - 3^x + 3 &=& 0, \\ 3^{x+2} \cdot 3^x - 3^{x+2} \cdot 3 - (3^x - 3) &=& 0, \\ (3^{x+2} - 1) (3^x - 3) &=& 0. \end{aligned}

Now that it is factored, we solve two equations separately. First,

3x+21=0,3x+2=1,93x=1,3x=1/9,x=2. \begin{aligned} 3^{x+2} - 1 &=& 0, \\ 3^{x+2} &=& 1, \\ 9 \cdot 3^x &=& 1, \\ 3^x &=& 1/9, \\ x &=& -2. \end{aligned}

Then,

3x3=0,3x=3,x=1. \begin{aligned} 3^x - 3 &=& 0, \\ 3^x &=& 3, \\ x &=& 1. \end{aligned}

We've found two roots.