Question

Solve ${2^{2x + 2}} - {6^x} - {2.3^{2x + 2}} = 0$

Answer 1

Hint: Whenever you see a question like this, just think of how we convert it into a quadratic equation. Here first we have to convert this equation in the form of ${2^x}$ and ${3^x}$. Then we have to divide the whole equation by either ${2^x}$ or ${3^x}$.

Complete step by step solution:
In the given question, we have
\[ \Rightarrow {2^{2x + 2}} - {6^x} - {2.3^{2x + 2}} = 0\]
Now, we can also write it as
\[ \Rightarrow {2^{2\left( {x + 1} \right)}} - {6^x} - {2.3^{2\left( {x + 1} \right)}} = 0\]
\[ \Rightarrow {4^{\left( {x + 1} \right)}} - {6^x} - {2.9^{\left( {x + 1} \right)}} = 0\]
We know that ${a^{m + n}} = {a^m}.{a^n}$ and ${4^x} = {\left( {{2^x}} \right)^2}$
Also, ${\left( {a \times b} \right)^x} = {a^x}.{b^x}$
\[ \Rightarrow 4.{\left( {{2^x}} \right)^2} - {2^x}{.3^x} - 2.\left( {{9^x}.9} \right) = 0\]
On multiplying, we get
\[ \Rightarrow 4.{\left( {{2^x}} \right)^2} - {2^x}{.3^x} - 18{\left( {{3^x}} \right)^2} = 0\]
On dividing with ${\left( {{2^x}} \right)^2}$, we get
$ \Rightarrow \dfrac{{4.{{\left( {{2^x}} \right)}^2}}}{{{{\left( {{2^x}} \right)}^2}}} - \dfrac{{{2^x}{{.3}^x}}}{{{{\left( {{2^x}} \right)}^2}}} - 18{\left[ {{{\left( {\dfrac{3}{2}} \right)}^x}} \right]^2} = 0$
$ \Rightarrow 4 - {\left( {\dfrac{3}{2}} \right)^x} - 18{\left[ {{{\left( {\dfrac{3}{2}} \right)}^x}} \right]^2} = 0$
Let, ${\left( {\dfrac{3}{2}} \right)^x} = y$
Now,
$ \Rightarrow 4 - y - 18{y^2} = 0$
On factorization, we get
$ \Rightarrow 4 - \left( {9 - 8} \right)y - 18{y^2} = 0$
$ \Rightarrow 4 - 9y + 8y - 18{y^2} = 0$
$ \Rightarrow 1\left( {4 - 9y} \right) + 2y\left( {4 - 9y} \right) = 0$
$ \Rightarrow \left( {4 - 9y} \right)\left( {1 + 2y} \right) = 0$
Now, there are two cases
Case1: $4 - 9y = 0$
$ \Rightarrow y = \dfrac{4}{9}$
Case 2: $1 + 2y = 0$
$ \Rightarrow y = - \dfrac{1}{2}$
So, we have to neglect case $2$ because the value of y can never be negative. It is so because the power of any power of any positive number gives a positive value.
Therefore
$ \Rightarrow y = \dfrac{4}{9}$
Now, put the value of $y = \,{\left( {\dfrac{3}{2}} \right)^x}$
Therefore,
$ \Rightarrow {\left( {\dfrac{3}{2}} \right)^x} = \dfrac{4}{9}$
$ \Rightarrow {\left( {\dfrac{3}{2}} \right)^x} = {\left( {\dfrac{2}{3}} \right)^2}$
$ \Rightarrow {\left( {\dfrac{3}{2}} \right)^x} = {\left( {\dfrac{3}{2}} \right)^{ - 2}}$
On comparing the power,
$x = - 2$
Therefore, the value of x is $ - 2$ .
So, the correct answer is “-2”.

Note: There can be many variations in these types of questions. As in this question we need to divide it by ${\left( {{2^x}} \right)^2}$. But it is not always required. There can be many straightforward questions where there is no need for such division. We just need to identify which term we have to use as a variable. Also, one more thing to be noted is that we should take only those values of assumed variables which are positive. Negative values should be neglected.