Question

How do you solve \[{{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{32} \right)}^{3x+1}}\]?

Answer 1

Hint: This type of problem is based on the concept of equating the powers when the base is the same. First, we have to consider the LHS. We can write 4 as \[{{2}^{2}}\]. Using the property of powers, that is \[{{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}\], we get the LHS as \[{{\left( \dfrac{1}{2} \right)}^{4x}}\]. Now we need to consider the RHS. We can write 32 as \[{{2}^{5}}\]. Using the property \[{{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}\], we get RHS as \[{{\left( \dfrac{1}{2} \right)}^{5\left( 3x+1 \right)}}\]. We know that, if \[{{a}^{m}}={{a}^{n}}\], then m=n. with this rule, we get 4x=5(3x+1). Use distributive property and simplify the RHS. Do necessary calculations to find the value of x.

Complete step by step solution:
According to the question, we are asked to solve the equation \[{{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{32} \right)}^{3x+1}}\].
We have been given the inequality is \[{{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{32} \right)}^{3x+1}}\]. -----(1)
We first have to consider the LHS of the equation (1).
We know that, \[4={{2}^{2}}\].
We get \[{{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{{{2}^{2}}} \right)}^{2x}}\].
We also know that 1 to the power any term is x. thus, we can write 1 as \[{{1}^{2}}\].
\[\Rightarrow {{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{{{1}^{2}}}{{{2}^{2}}} \right)}^{2x}}\]
Using the division rule of powers, that is \[\dfrac{{{a}^{n}}}{{{b}^{n}}}={{\left( \dfrac{a}{b} \right)}^{n}}\]. We get
\[\Rightarrow {{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( {{\left( \dfrac{1}{2} \right)}^{2}} \right)}^{2x}}\].
But, we know that \[{{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}\]. Using this property, we get
\[{{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{2} \right)}^{2\times 2x}}\]
On further simplification, we get
\[{{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{2} \right)}^{4x}}\] ------------(2)
Now, let us consider that RHS.
We know that, \[32={{2}^{5}}\].
We get \[{{\left( \dfrac{1}{32} \right)}^{3x+1}}={{\left( \dfrac{1}{{{2}^{5}}} \right)}^{3x+1}}\].
We also know that 1 to the power any term is x. thus, we can write 1 as \[{{1}^{5}}\].
\[\Rightarrow {{\left( \dfrac{1}{32} \right)}^{3x+1}}={{\left( \dfrac{{{1}^{5}}}{{{2}^{5}}} \right)}^{3x+1}}\]
Using the division rule of powers, that is \[\dfrac{{{a}^{n}}}{{{b}^{n}}}={{\left( \dfrac{a}{b} \right)}^{n}}\]. We get
\[\Rightarrow {{\left( \dfrac{1}{32} \right)}^{3x+1}}={{\left( {{\left( \dfrac{1}{2} \right)}^{5}} \right)}^{3x+1}}\].
But, we know that \[{{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}\]. Using this property, we get
\[{{\left( \dfrac{1}{32} \right)}^{3x+1}}={{\left( \dfrac{1}{2} \right)}^{5\left( 3x+1 \right)}}\]------------(3)
Let us now equate the simplified LHS and RHS. We get
\[{{\left( \dfrac{1}{2} \right)}^{4x}}={{\left( \dfrac{1}{2} \right)}^{5\left( 3x+1 \right)}}\]
Here, we find that the base is the same that is \[\dfrac{1}{2}\].
We know that, if \[{{a}^{m}}={{a}^{n}}\], then m=n. using this rule in the above equation, we get
\[4x=5\left( 3x+1 \right)\] ----------(4)
Let us now use distributive property, that is \[a\left( b+c \right)=ab+ac\], in the above expression.
We get
\[4x=5\times 3x+1\times 5\]
\[\Rightarrow 4x=5\times 3x+5\]
On further simplification, we get
\[4x=15x+5\]
Let us now subtract 15x from both the sides of the equation.
\[\Rightarrow 4x-15x=15x+5-15x\]
We know that terms with the same magnitude and opposite signs cancel out.
Therefore, on cancelling 15x from the RHS, we get
\[4x-15x=5\]
Now, let us group the x terms and add them.
\[\Rightarrow \left( 4-15 \right)x=5\]
On further simplification, we get
-11x=5.
Now, we have to divide the whole equation by -11.
\[\Rightarrow \dfrac{-11x}{-11}=\dfrac{5}{-11}\]
We find that -11 are common in both the numerator and denominator of the LHS. On cancelling -11, we get
\[x=\dfrac{5}{-11}\]
\[\therefore x=\dfrac{-5}{11}\]

Therefore, the value of x in the equation \[{{\left( \dfrac{1}{4} \right)}^{2x}}={{\left( \dfrac{1}{32} \right)}^{3x+1}}\] is \[\dfrac{-5}{11}\].

Note: We should know the properties and rules of powers to solve this type of questions. We should avoid calculation mistakes based on sign conventions. We should separately solve the linear equation with variable x by equating the powers to get the value of x.