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How do you solve for x in ${{3}^{2x}}+{{3}^{x+1}}-4=0$?

Last updated date: 09th Aug 2024
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Hint: Use the formula ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ and ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$ to simplify the L.H.S. of the given expression. Now, assume ${{3}^{x}}=k$ and form a quadratic equation in k. Use the middle term split method to solve for the value of k. Reject the negative value of k by using the fact that ‘exponential function cannot be negative’. Take log to the base 3 both the sides to get the answer.

Complete step-by-step solution:
Here, we have been provided with the equation: - ${{3}^{2x}}+{{3}^{x+1}}-4=0$ and we are asked to determine the value of x.
$\because {{3}^{2x}}+{{3}^{x+1}}-4=0$
Using the formula: - ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ and ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$, we get,
\begin{align} & \Rightarrow {{\left( {{3}^{x}} \right)}^{2}}+{{3}^{x}}\times {{3}^{1}}-4=0 \\ & \Rightarrow {{\left( {{3}^{x}} \right)}^{2}}+{{3}^{x}}.3-4=0 \\ \end{align}
Assuming ${{3}^{x}}=k$, we get,
$\Rightarrow {{k}^{2}}+3k-4=0$
Using the middle term split method, we get,
\begin{align} & \Rightarrow {{k}^{2}}+4k-k-4=0 \\ & \Rightarrow k\left( k+4 \right)-1\left( k+4 \right)=0 \\ \end{align}
Taking $\left( k+4 \right)$ common, we get,
$\Rightarrow \left( k+4 \right)\left( k-1 \right)=0$
Substituting each term equal to 0, we get,
$\Rightarrow k+4=0$ or $k-1=0$
$\Rightarrow k=-4$ or $k=1$
$\Rightarrow {{3}^{x}}=-4$ or ${{3}^{x}}=1$
Now, we have obtained two values of ${{3}^{x}}$, let us check if any of the two values is invalid or not. Since, we know that ${{3}^{x}}$ is an exponential function and the value of an exponential function is always positive for any value of the variable x. In the above two values of ${{3}^{x}}$ we have one negative value, i.e. -4, which cannot be possible. Therefore ${{3}^{x}}=-4$ must be rejected.
$\Rightarrow {{3}^{x}}=1$
Taking log to the base 3 both the sides, we get,
$\Rightarrow {{\log }_{3}}{{3}^{x}}={{\log }_{3}}1$
Using the formula: - $\log {{a}^{m}}=m\log a$ and $\log 1=0$, we get,
\begin{align} & \Rightarrow x{{\log }_{3}}3=0 \\ & \Rightarrow x=0 \\ \end{align}
Hence, the value of x is 0.

Note: One may check the answer by substituting the value x = 0 in the original equation. If L.H.S. = R.H.S. then our answer is correct. You must know the graph of an exponential function and remember that it is never negative for any value of the variable. It is necessary to check the obtained values of k and reject the invalid value otherwise the answer will be considered incorrect. Note that here we have used the middle term split method to get the value of k, you can use the discriminant method and get the answer. Here, $k=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where a = coefficient of ${{k}^{2}}$, b = coefficient of k and c = constant term. Remember that the value of k is not our answer but we need to find the value of x.