Question

How do you solve the equation ${{2}^{3x+1}}={{3}^{x-2}}$ .

Answer 1

Hint: Now to solve the equation we will first simplify the equation by using properties of indices. We will try to write the equation in the form ${{\left( \dfrac{m}{n} \right)}^{x}}=\dfrac{p}{q}$ . Once we have the equation in this form we will take logs on both sides. Then we will use the properties of log to simplify the equation and hence find the value of x.

Complete step-by-step answer:
Now we know that ${{x}^{m+n}}={{x}^{m}}{{x}^{n}}$ and ${{x}^{m-n}}=\dfrac{{{x}^{m}}}{{{x}^{n}}}$ hence using this we can rewrite the equation as
$\Rightarrow {{2}^{3x}}.2=\dfrac{{{3}^{x}}}{{{3}^{2}}}$
Now dividing the whole equation by 2 we get,
$\Rightarrow {{2}^{3x}}=\dfrac{{{3}^{x}}}{{{2.3}^{2}}}$
Now we know that ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$ . Hence using this property we get,
 $\begin{align}
  & \Rightarrow {{\left( {{2}^{3}} \right)}^{x}}=\dfrac{{{3}^{x}}}{18} \\
 & \Rightarrow {{8}^{x}}=\dfrac{{{3}^{x}}}{18} \\
\end{align}$
Now taking ${{3}^{x}}$ on RHS we get,
$\Rightarrow \dfrac{{{8}^{x}}}{{{3}^{x}}}=\dfrac{1}{18}$
$\Rightarrow {{\left( \dfrac{8}{3} \right)}^{x}}=\dfrac{1}{18}$
Now to solve this equation we will take log on both sides
Hence we get,
$\Rightarrow \log {{\left( \dfrac{8}{3} \right)}^{x}}=\log \left( \dfrac{1}{18} \right)$
Now we know the property of log which says $\log {{a}^{m}}=m\log a$ hence using this property we get,
$\Rightarrow x\log \left( \dfrac{8}{3} \right)=\log \left( \dfrac{1}{18} \right)$
Now we know that $\log \left( \dfrac{a}{b} \right)=\log a-\log b$ and hence we get
$\begin{align}
  & \Rightarrow x\left[ \log 8-\log 3 \right]=\log 1-\log 18 \\
 & \Rightarrow x\left[ \log {{2}^{3}}-\log 3 \right]=0-\log \left( 9\times 2 \right) \\
\end{align}$Now again using $\log {{a}^{m}}=m\log a$ and $\log \left( ab \right)=\log a+\log b$ we get,
$\begin{align}
  & \Rightarrow x\left[ 3\times \log 2-\log 3 \right]=-\log 9-\log 2 \\
 & \Rightarrow x\left[ 3\times \log 2-\log 3 \right]=-\log {{3}^{2}}-\log 2 \\
 & \Rightarrow x\left[ 3\times \log 2-\log 3 \right]=-2\times \log 3-\log 2 \\
\end{align}$Now we know that log2 = 0.3 and log 3 = 0.477
Now substituting the values in the equation we get,
$\begin{align}
  & \Rightarrow x\left[ 3\times 0.3-0.477 \right]=-2\times 0.477-0.3 \\
 & \Rightarrow 0.423x=1.254 \\
 & \Rightarrow x=\dfrac{1.254}{0.423}=2.96 \\
\end{align}$
Hence the solution of the equation is 2.96.

Note: Note that there will not be an integer solution of the equation as $2\times 2\times 2...$ will be even while $3\times 3\times 3...$ will always be odd. Hence x will never be an integer but will be a rational number. Now not that if here we had the variable in power. Whenever we have variable in power we can take log and simplify the equation. We can also directly take log and solve the equation but for simple calculations we have first simplified the equation and then used log to solve the equation.