Concept
If the bases of exponential terms are same then their powers will also be equal. If [tex]a^x=a^y[/tex] then [tex]x=y[/tex].
Given
[tex]7^{-2} \times 3^2=(\dfrac{3}{7} )^{4x}[/tex]
To find
Value of [tex]x[/tex]
Solution
It is given that
[tex]\Rightarrow\dfrac{1}{7^2} \times 3^2=\left ( \frac{3}{7} \right )^{4x}\\\Rightarrow \left ( \dfrac{3}{7} \right )^2= \left ( \frac{3}{7} \right )^{4x}[/tex]
Since the bases [tex]\dfrac{3}{7}[/tex] are equal, so their powers will also be equal
[tex]\Rightarrow 2=4x\\\Rightarrow x=\dfrac{1}{2}[/tex]
As a result, value of [tex]x[/tex] is [tex]\dfrac{1}{2}[/tex]
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Answers & Comments
Concept
If the bases of exponential terms are same then their powers will also be equal. If [tex]a^x=a^y[/tex] then [tex]x=y[/tex].
Given
[tex]7^{-2} \times 3^2=(\dfrac{3}{7} )^{4x}[/tex]
To find
Value of [tex]x[/tex]
Solution
It is given that
[tex]7^{-2} \times 3^2=(\dfrac{3}{7} )^{4x}[/tex]
[tex]\Rightarrow\dfrac{1}{7^2} \times 3^2=\left ( \frac{3}{7} \right )^{4x}\\\Rightarrow \left ( \dfrac{3}{7} \right )^2= \left ( \frac{3}{7} \right )^{4x}[/tex]
Since the bases [tex]\dfrac{3}{7}[/tex] are equal, so their powers will also be equal
[tex]\Rightarrow 2=4x\\\Rightarrow x=\dfrac{1}{2}[/tex]
As a result, value of [tex]x[/tex] is [tex]\dfrac{1}{2}[/tex]