Courses
Courses for Kids
Free study material
Offline Centres
More

If \[{{3}^{x+1}}={{6}^{{{\log }_{2}}3}}\] then ‘\[x\]’ is

seo-qna
Last updated date: 22nd Feb 2024
Total views: 338.4k
Views today: 4.38k
IVSAT 2024
Answer
VerifiedVerified
338.4k+ views
Hint: By using exponential formulae, \[{{\left( a.b \right)}^{m}}={{a}^{m}}.{{b}^{m}}\] we should split the R.H.S term. Now as we know the formula \[{{a}^{{{\log }_{a}}b=}}b\] . So by using this formula, we should evaluate the split term of R.H.S. We know the formula \[\text{if }{{a}^{x}}={{a}^{n}}\]then \[x=n\]. Now by using the formula, we can find the value of x.

Complete step by step answer:
 We are given to solve this question we are given to find \[x\]for the equation \[{{3}^{x+1}}={{6}^{{{\log }_{2}}3}}\] .
Let us consider the given equation as equation (1).
\[{{3}^{x+1}}={{6}^{{{\log }_{2}}3}}...................\left( 1 \right)\]
Now we have to find x from the equation (1). For that we have to bring the ‘x’ from exponent to the base.
Let us rewrite the equation (1) by writing \[6\] as \[{{3}^{2}}\], we get
\[\Rightarrow {{3}^{x+1}}={{\left( 3.2 \right)}^{{{\log }_{2}}3}}\]
Let us consider the above equation as equation (2).
\[{{3}^{x+1}}={{\left( 3.2 \right)}^{{{\log }_{2}}3}}.................\left( 2 \right)\]
As we know the formula \[{{\left( a.b \right)}^{m}}={{a}^{m}}.{{b}^{m}}\]
Let us consider the formula as (f1).
\[{{\left( a.b \right)}^{m}}={{a}^{m}}.{{b}^{m}}.......................\left( f1 \right)\]
Now let us apply the formula (f1) to the equation (2), we get
\[\Rightarrow {{3}^{x+1}}={{\left( 3 \right)}^{{{\log }_{2}}3}}.{{\left( 2 \right)}^{{{\log }_{2}}3}}\]
Let us consider the above equation as equation (3).
\[\Rightarrow {{3}^{x+1}}={{\left( 3 \right)}^{{{\log }_{2}}3}}.{{\left( 2 \right)}^{{{\log }_{2}}3}}.................\left( 3 \right)\]
Now as we know the formula \[{{a}^{{{\log }_{a}}b=}}b\] .Let we consider the formula as (f2).
\[{{a}^{{{\log }_{a}}b=}}b.................\left( f2 \right)\]
By applying the formula (f2) to the equation (3), we get
\[\Rightarrow {{3}^{x+1}}={{\left( 3 \right)}^{{{\log }_{2}}3}}.3\]
Now let us divide the above equation by 3 on both sides, we get
\[\Rightarrow \dfrac{{{3}^{x+1}}}{3}={{\left( 3 \right)}^{{{\log }_{2}}3}}\]
Now by simplifying a bit to the above equation, we get
\[\Rightarrow {{3}^{x}}={{\left( 3 \right)}^{{{\log }_{2}}3}}\]
Let us consider the above equation as equation (4), we get
\[\Rightarrow {{3}^{x}}={{3}^{{{\log }_{2}}3}}\]
Let us consider the above equation as equation (4).
\[\Rightarrow {{3}^{x}}={{3}^{{{\log }_{2}}3}}...............\left( 4 \right)\]
As we know the formula \[\text{if }{{a}^{x}}={{a}^{n}}\]then \[x=n\].
Let us apply above concept to the equation (4), we get
\[\Rightarrow x={{\log }_{2}}3\]
Let us consider the above equation as equation (5).
\[\Rightarrow x={{\log }_{2}}3.............\left( 5 \right)\]
Therefore equation (5) is the solution for a given problem.

Note: Students may have a misconception that \[{{a}^{{{\log }_{a}}b=}}a\] but we know that \[{{a}^{{{\log }_{a}}b=}}b\]. So, if this misconception is followed the final answer may get interrupted. So, these misconceptions should be avoided while solving the problem. Hence, students should have a clear idea of the respective concepts.