How do you solve \[{{3}^{x+1}}+{{3}^{x}}=36\] ?
Answer
625.5k+ views
Hint: The given problem and its various types are very easy to solve as well as pretty simple to understand. Before solving the problem, we need to keep in mind a few things about fraction and indices. We basically have to remember two concepts which goes like,
\[{{x}^{a+b}}={{x}^{a}}\cdot {{x}^{b}}\] and \[{{x}^{a-b}}=\dfrac{{{x}^{a}}}{{{x}^{b}}}\] where x ,a ,b are any variables or parameters.
Keeping in mind of these two concepts, we can therefore solve the given problem very easily.
Complete step by step answer:
Now, using the relation \[{{x}^{a+b}}={{x}^{a}}\cdot {{x}^{b}}\] , we apply it in our given problem, we therefore write,
\[\begin{align}
& {{3}^{x+1}}={{3}^{x}}\cdot {{3}^{1}} \\
& \Rightarrow {{3}^{x+1}}=3\cdot {{3}^{x}} \\
\end{align}\]
Now, applying it in the original problem we get,
\[\begin{align}
& {{3}^{x+1}}+{{3}^{x}}=36 \\
& \Rightarrow 3\cdot {{3}^{x}}+{{3}^{x}}=36 \\
\end{align}\]
Now, taking \[{{3}^{x}}\] as common, we get,
\[\begin{align}
& {{3}^{x}}\left( 3+1 \right)=36 \\
& \Rightarrow {{3}^{x}}\left( 4 \right)=36 \\
& \Rightarrow {{3}^{x}}=\dfrac{36}{4} \\
& \Rightarrow {{3}^{x}}=9 \\
\end{align}\]
Now, we need to express \[9\] as a power of \[3\] , which we can do very easily. We know that we can write \[9\] as \[3\times 3\] . Now following the same procedure we write, \[9=3\times 3\] and we know that \[3\] can simply be written as \[{{3}^{1}}\] . Thus we write \[9={{3}^{1}}\times {{3}^{1}}\] \[\Rightarrow 9={{3}^{2}}\] .
Thus we can say \[{{3}^{x}}={{3}^{2}}\] . Now, comparing both the sides of the equation, we can write,
\[x=2\]
So the answer to this problem is \[x=2\] . We can cross check the answer by putting this value of \[x\] in the original given equation, from which we can easily find that both the L.H.S and R.H.S match, which is a proof that we have solved the question correctly.
Note: In these types of sums we need to remember the basic equations of exponent and indices and then solve the problems by a simple linear method as shown above. The common mistake that students can make here they would overlook the basic properties of exponents and write \[{{3}^{x+1}}\] as \[{{3}^{x}}+{{3}^{1}}\] . this would result in wrong answers.
\[{{x}^{a+b}}={{x}^{a}}\cdot {{x}^{b}}\] and \[{{x}^{a-b}}=\dfrac{{{x}^{a}}}{{{x}^{b}}}\] where x ,a ,b are any variables or parameters.
Keeping in mind of these two concepts, we can therefore solve the given problem very easily.
Complete step by step answer:
Now, using the relation \[{{x}^{a+b}}={{x}^{a}}\cdot {{x}^{b}}\] , we apply it in our given problem, we therefore write,
\[\begin{align}
& {{3}^{x+1}}={{3}^{x}}\cdot {{3}^{1}} \\
& \Rightarrow {{3}^{x+1}}=3\cdot {{3}^{x}} \\
\end{align}\]
Now, applying it in the original problem we get,
\[\begin{align}
& {{3}^{x+1}}+{{3}^{x}}=36 \\
& \Rightarrow 3\cdot {{3}^{x}}+{{3}^{x}}=36 \\
\end{align}\]
Now, taking \[{{3}^{x}}\] as common, we get,
\[\begin{align}
& {{3}^{x}}\left( 3+1 \right)=36 \\
& \Rightarrow {{3}^{x}}\left( 4 \right)=36 \\
& \Rightarrow {{3}^{x}}=\dfrac{36}{4} \\
& \Rightarrow {{3}^{x}}=9 \\
\end{align}\]
Now, we need to express \[9\] as a power of \[3\] , which we can do very easily. We know that we can write \[9\] as \[3\times 3\] . Now following the same procedure we write, \[9=3\times 3\] and we know that \[3\] can simply be written as \[{{3}^{1}}\] . Thus we write \[9={{3}^{1}}\times {{3}^{1}}\] \[\Rightarrow 9={{3}^{2}}\] .
Thus we can say \[{{3}^{x}}={{3}^{2}}\] . Now, comparing both the sides of the equation, we can write,
\[x=2\]
So the answer to this problem is \[x=2\] . We can cross check the answer by putting this value of \[x\] in the original given equation, from which we can easily find that both the L.H.S and R.H.S match, which is a proof that we have solved the question correctly.
Note: In these types of sums we need to remember the basic equations of exponent and indices and then solve the problems by a simple linear method as shown above. The common mistake that students can make here they would overlook the basic properties of exponents and write \[{{3}^{x+1}}\] as \[{{3}^{x}}+{{3}^{1}}\] . this would result in wrong answers.
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