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

Last updated date: 22nd Feb 2024
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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.

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.