Question

Solution of the equation \[4 \times {9^{x - 1}}\; = {\text{ }}3\sqrt {({2^{2x + 1}})} \] is
A. \[3\]
B. \[2\]
C. \[3/2\]
D. \[2/3\]

Answer 1

Hint: In this question, we need to solve the equation \[4 \times {9^{x - 1}}\; = {\text{ }}3\sqrt {({2^{2x + 1}})} \]. For this, we have to simplify the equation \[4 \times {9^{x - 1}}\; = {\text{ }}3\sqrt {({2^{2x + 1}})} \] to find the value of x. We will use the exponential properties for simplifying this equation.

Formula used: The following exponential properties are used to solve this question.
\[{a^m} \times {a^n} = {a^{m + n}}\] and \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]

Complete step-by-step solution:
We know that \[4 \times {9^{x - 1}}\; = {\text{ }}3\sqrt {({2^{2x + 1}})} \]
Let us simplify the above mathematical equation.
Thus, we get
\[\dfrac{{{9^{x - 1}}}}{3}\; = {\text{ }}\dfrac{{\sqrt {({2^{2x + 1}})} }}{4}\]
But we know that \[{3^2} = 9\] and \[{2^2} = 4\]
So, we get
\[\dfrac{{{{\left( {{3^2}} \right)}^{x - 1}}}}{3}\; = {\text{ }}\dfrac{{\sqrt {({2^{2x + 1}})} }}{{{2^2}}}\]
But \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
Thus, we get
\[\dfrac{{{{\left( 3 \right)}^{2x - 2}}}}{{{3^1}}}\; = {\text{ }}\dfrac{{\sqrt {({2^{2x + 1}})} }}{{{2^2}}}\]
\[{\left( 3 \right)^{2x - 2 - 1}}\; = {\text{ }}\dfrac{{\sqrt {({2^{2x + 1}})} }}{{{2^2}}}\]
\[{\left( 3 \right)^{2x - 3}}\; = {\text{ }}\sqrt {\dfrac{{({2^{2x + 1}})}}{{16}}} \]
But \[{2^4} = 16\] and
So, we get
\[{\left( 3 \right)^{2x - 3}}\; = {\text{ }}\sqrt {\dfrac{{({2^{2x + 1}})}}{{{2^4}}}} \]
But \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
By applying exponential property we get
\[{\left( 3 \right)^{2x - 3}}\; = {\text{ }}\sqrt {({2^{2x + 1 - 4}})} \]
\[{\left( 3 \right)^{2x - 3}}\; = {\text{ }}\sqrt {({2^{2x - 3}})} \]
We can write \[\sqrt {({2^{2x - 3}})} \] as \[{({2^{2x - 3}})^{1/2}}\]
Thus, we get
\[{\left( 3 \right)^{2x - 3}}\; = {\text{ }}{({2^{2x - 3}})^{\dfrac{1}{2}}}\]
But we know that, \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\],
By using, \[{a^m} \times {a^n} = {a^{m + n}}\], we get

\[{\left( 3 \right)^{2x - 3}}\; = {\text{ }}{(2)^{\dfrac{{2x - 3}}{2}}}\]
By taking square on both sides, we get
\[{\left( 3 \right)^{2 \times \left( {2x - 3} \right)}}\; = {\text{ }}{(2)^{\dfrac{{2x - 3}}{2} \times 2}}\]
\[{\left( 3 \right)^{\left( {4x - 6} \right)}}\; = {\text{ }}{(2)^{2x - 3}}\]
Here, we can say that
\[4x - 6 = 0\] or \[2x - 3 = 0\]
Thus, by simplifying, we get
\[4x = 6\]
\[x = 3/2\]
Also, we get
\[2x - 3 = 0\]
That is \[x = 3/2\].
Hence, the solution of the mathematical equation \[4 \times {9^{x - 1}}\; = {\text{ }}3\sqrt {({2^{2x + 1}})} \] is \[x = 3/2\].
Therefore, the correct option is (C).

Additional information: In mathematics, an expression is a statement that includes at least two numbers/variables as well as at least one math operation. So, we can say that the expressions in mathematics are mathematical declarations that include at least two terms that comprise numbers, variables, or both, and are linked by an operator in between. Addition, subtraction, multiplication, and division are examples of mathematical operators. Exponential expressions are simply a shorthand way of writing powers. The exponent shows how many times the base has been used as a factor.

Note: Many students generally make mistakes in the simplification part of \[4 \times {9^{x - 1}}\; = {\text{ }}3\sqrt {({2^{2x + 1}})} \]. They may get wrong while writing the exponential properties specifically while writing their signs. If so, then they may get end result wrong.