Question

How can I find x? (Exponential Equation)
${{2}^{3x+1}}={{3}^{x-2}}$

Answer 1

Hint: To solve the given equation, we need to use the properties of the exponents so as to separate x on the LHS. For this we have to use the property of the summation of exponents, given by ${{a}^{m+n}}={{a}^{m}}{{a}^{n}}$, the property of negative exponents, given by ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$, and the property of the multiplication of the exponents, given by $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$. Then, after separating x on the LHS, we have to take the logarithm on both the sides of the obtained equation which can be easily solved for x using the logarithm property $\log \left( {{a}^{m}} \right)=m\log a$.

Complete step by step solution:
The given equation in the above question is
$\Rightarrow {{2}^{3x+1}}={{3}^{x-2}}$
From the properties of the exponents, we know that ${{a}^{m+n}}={{a}^{m}}{{a}^{n}}$. Applying this on the LHS and RHS of the above equation, we can write the above equation as
$\begin{align}
  & \Rightarrow {{2}^{3x}}{{2}^{1}}={{3}^{x}}{{3}^{-2}} \\
 & \Rightarrow 2\left( {{2}^{3x}} \right)={{3}^{x}}{{3}^{-2}} \\
\end{align}$
Now, from the negative exponent property we know that ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$. Applying this on the RHS, we get
\[\begin{align}
  & \Rightarrow 2\left( {{2}^{3x}} \right)=\dfrac{{{3}^{x}}}{{{3}^{2}}} \\
 & \Rightarrow 2\left( {{2}^{3x}} \right)=\dfrac{{{3}^{x}}}{9} \\
\end{align}\]
Multiplying both the sides of the above equation by $9$, we get
\[\begin{align}
  & \Rightarrow 9\left[ 2\left( {{2}^{3x}} \right) \right]=9\left( \dfrac{{{3}^{x}}}{9} \right) \\
 & \Rightarrow 18\left( {{2}^{3x}} \right)={{3}^{x}} \\
\end{align}\]
From the properties of the products of exponents, we know that ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$. Applying this on the LHS of the above equation, we get
\[\begin{align}
  & \Rightarrow 18{{\left( {{2}^{3}} \right)}^{x}}={{3}^{x}} \\
 & \Rightarrow 18\left( {{8}^{x}} \right)={{3}^{x}} \\
\end{align}\]
Dividing both the sides of the above equation by \[{{8}^{x}}\], we get
\[\begin{align}
  & \Rightarrow \dfrac{18\left( {{8}^{x}} \right)}{{{8}^{x}}}=\dfrac{{{3}^{x}}}{{{8}^{x}}} \\
 & \Rightarrow 18=\dfrac{{{3}^{x}}}{{{8}^{x}}} \\
 & \Rightarrow \dfrac{{{3}^{x}}}{{{8}^{x}}}=18 \\
\end{align}\]
Applying the exponent property $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$ on the LHS of the above equation, we get
$\Rightarrow {{\left( \dfrac{3}{8} \right)}^{x}}=18$
Taking logarithm on both the sides, we get
$\Rightarrow \log \left[ {{\left( \dfrac{3}{8} \right)}^{x}} \right]=\log \left( 18 \right)$
From the properties of the logarithm function, we know that $\log \left( {{a}^{m}} \right)=m\log a$. Applying this on the LHS of the above equation, we get
$\Rightarrow x\log \left( \dfrac{3}{8} \right)=\log \left( 18 \right)$
Finally, dividing both sides of the above equation by $\log \left( \dfrac{3}{8} \right)$, we get
$\Rightarrow x=\dfrac{\log \left( 18 \right)}{\log \left( \dfrac{3}{8} \right)}$
Hence, the value of x is found to be equal to $\dfrac{\log \left( 18 \right)}{\log \left( \dfrac{3}{8} \right)}$.

Note: For solving such types of questions, we must be familiar with all the properties of the exponential and the logarithmic functions. The given equation may look difficult due to the different bases on the LHS and the RHS of the given equation. But after taking the logarithm it will get simplified, as in the above solution.