Question

Solve each exponential equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. Verify your results using a graphing utility.

$3·4^x+4·2^x+8=0$

Short answer

The given equation has no real solution.

Step-by-step solution

Step 1. Given information.

The given equation is:

$3·4^x+4·2^x+8=0$

It can be rewritten as:

$\begin{array}{rcl}3·(2^2{)}^x+4·2^x+8& =& 0\\ 3(2^x{)}^2+4(2^x)+8& =& 0\end{array}$

Step 2. Substitute another variable and solve for it.

Let $2^x=u$.

$3u^2+4u+8=0$

The discriminant of the above equation is:

$\begin{array}{rcl}D& =& b^2-4ac\\ & =& (4{)}^2-4(3\left)\right(8)\\ & =& 16-96\\ & =& -80\end{array}$

The discriminant is negative. It means the equation has no real solution.

Step 3. Verify the result by the graphing utility.

Plot the graph of $y=3·4^x+4·2^x+8$on a coordinate plane. The x-intercept of the curve is the solution of the given equation.

The graph does not intersect the x-axis at any point. So, the given equation has no real solution. Hence the solution is verified.