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How do you solve \[243={{9}^{2X+1}}\]?

seo-qna
Last updated date: 15th Jul 2024
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Answer
VerifiedVerified
381.9k+ views
Hint: By using the transformations, we can make the given question very easy to do. From the question given we have been asked to solve \[243={{9}^{2X+1}}\]. We know that we can express 243 as well as 9 as a power of 3. Then, we will make use of formula \[\Rightarrow {{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\] and simplify further to get the value of x.

Complete step-by-step solution:
From the question it had been given that \[243={{9}^{2X+1}}\]
\[\Rightarrow 243={{9}^{2X+1}}\]
We know that \[243\] can be written as \[{{3}^{5}}\]
By substituting it in the given question, we get the below equation,
\[\Rightarrow {{3}^{5}}={{9}^{2X+1}}\]
Similarly, we know that \[9\] can be written as \[{{3}^{2}}\]
Now, we know the basic formula of exponents,
\[\Rightarrow {{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\]
By substituting it in the given question, we get the below equation, and
By using the above basic formula of exponents, we get,
\[\Rightarrow {{3}^{5}}={{3}^{2\left( 2x+1 \right)}}\]
As the bases are equal in the above equation, we can equate the indices or powers.
By equating the indices or powers of the above equation, we get,
\[\Rightarrow 5=2\left( 2x+1 \right)\]
Furthermore simplifying the above equation we get,
\[\Rightarrow 5=4x+2\]
Now, take of\[2\] from both sides we get,
\[\Rightarrow 5-2=4x+2-2\]
\[\Rightarrow 3=4x\]
Now, divide both sides with \[4\] we get,
\[\Rightarrow x=\dfrac{3}{4}\]
Hence, the given equation is simplified.

Note: Students should be well aware of the exponents and powers. Students should be very careful while doing the calculation we should not make mistakes like here in this question \[\Rightarrow 243={{9}^{2x+1}}\] we should not equate \[\Rightarrow 5=2x+1\] like this we should equate it to \[\Rightarrow 5=4x+2\] to get answer in this type of problems as calculation part is somewhat difficult in this type of problems. Students should know which type of transformation is to be used for the given question to get the question into an easier way.