Question

Solve each exponential equation. Express irrational solutions in exact form. $$ 3^{2 x}+3^{x}-2=0 $$

Short answer

The solution is \(x = 0\).

Step-by-step solution

Substitute the variable

Let’s make the substitution to turn the equation into a quadratic form. We set \(y = 3^x\). So, if \(y = 3^x\), then \(y^2 = (3^x)^2 = 3^{2x}\). The equation \(3^{2x} + 3^x - 2 = 0\) now becomes \(y^2 + y - 2 = 0\).

Solve the quadratic equation

Now we solve the quadratic equation \(y^2 + y - 2 = 0\). We can factorize it to get:\[(y + 2)(y - 1) = 0\].

Find the solutions for y

Setting each factor to zero gives the solutions for \(y\): \(y + 2 = 0\) results in \(y = -2\), and \(y - 1 = 0\) results in \(y = 1\).

Back-substitute to find x

Replace \(y\) with \(3^x\) to find the values of \(x\). Since \(y = -2\) is not possible because \(3^x\) is always positive, we discard this solution. So, we only consider \(3^x = 1\). Solving for \(x\), we get:\(3^x = 1\).This can be written as \(3^x = 3^0\), hence \(x = 0\).

Key concepts

quadratic substitution

When solving complex exponential equations, a helpful technique is to use quadratic substitution. This involves changing the form of the equation so it resembles a quadratic equation, which is easier to solve.
In our given problem, the equation is: \(3^{2x} + 3^{x} - 2 = 0\). By setting \(y = 3^{x}\), the equation can be rewritten as: \(y^2 + y - 2 = 0\).
This substitution simplifies the problem because we now deal with a familiar quadratic form. This strategic substitution transforms the exponential equation into a solvable quadratic equation. Quadratic substitution is particularly handy when the exponents are multiples of one another since it converts an otherwise complex problem into a straightforward one.

solving quadratic equations

Now that we have a quadratic equation \(y^2 + y - 2 = 0\), the next step is solving it. Quadratic equations can be solved by various methods, including factoring, completing the square, or using the quadratic formula.
In this instance, factoring works well. We look for two numbers that multiply to -2 and add to +1. A bit of thinking reveals that these numbers are +2 and -1.
Therefore, the quadratic equation can be factored as \((y + 2)(y - 1) = 0)\). This gives us two potential solutions: \(y + 2 = 0\) and \(y - 1 = 0 \).
Solving these, we find \(y = -2\) and \(y = 1\). These are the solutions in terms of \(y\). It is important to consistently check the feasibility of these solutions when they are substituted back into the original variable.

back-substitution

The final step involves back-substitution, where we replace \(y\) back with \(3^x\). This step is crucial as it translates our solutions from the quadratic back to the exponential form.
From the previous step, we have \(y = -2\) and \(y = 1\). Substituting back, we get \(3^x = -2\) and \(3^x = 1\).
Since exponential functions like \(3^x\) can never be negative, we discard \(3^x = -2\). This leaves us with \(3^x = 1\), which is true when \(x = 0\) since \(3^0 = 1\).
This method ensures that we accurately solve the original exponential equation while verifying the possible values of \(x\) through back-substitution. Always verify if the solutions are valid for the original equation after performing back-substitution.