Question

Solve: - ${{2}^{x+2}}\times {{27}^{\dfrac{x}{\left( x-1 \right)}}}=9$.
(a) $x=1-\dfrac{\log 3}{\log 2}$
(b) $x=-2$
(c) $x=1-\dfrac{\log 2}{\log 3}$
(d) $x=2$

Answer 1

Hint: First of all write 27 and 9 in their exponential form as${{3}^{3}}$ and ${{3}^{2}}$ respectively and use the formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to simplify. Now, take common log, i.e. log to the base 10, both the sides and use the properties of log given as: - $\log \left( m\times n \right)=\log m+\log n$, $\log \left( {{a}^{m}} \right)=m\log \left( a \right)$ to simplify. Shift all the terms to the L.H.S and take common terms together and write them as the product of two terms. Substitute each term equal to 0 and find the values of x.

Complete step by step answer:
Here we have been provided with the expression ${{2}^{x+2}}\times {{27}^{\dfrac{x}{\left( x-1 \right)}}}=9$ and we are asked to solve it. That means we have to find the value of x. Let us use some properties of exponents and logarithm to get the answer.
Now, converting the base 27 and 9 into their exponential form we have ${{3}^{3}}$ and ${{3}^{2}}$ respectively, so we get,
\[\Rightarrow {{2}^{x+2}}\times {{\left( {{3}^{3}} \right)}^{\dfrac{x}{\left( x-1 \right)}}}={{3}^{2}}\]
Using the formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ we get,
\[\Rightarrow {{2}^{x+2}}\times {{3}^{\dfrac{3x}{\left( x-1 \right)}}}={{3}^{2}}\]
Taking log to the base 10 both the sides and using the formulas $\log \left( m\times n \right)=\log m+\log n$, $\log \left( {{a}^{m}} \right)=m\log \left( a \right)$ to simplify, we get,
\[\begin{align}
  & \Rightarrow \log \left[ {{2}^{x+2}}\times {{3}^{\dfrac{3x}{\left( x-1 \right)}}} \right]=\log \left( {{3}^{2}} \right) \\
 & \Rightarrow \log \left[ {{2}^{x+2}} \right]+\log \left[ {{3}^{\dfrac{3x}{\left( x-1 \right)}}} \right]=2\log 3 \\
 & \Rightarrow \left( x+2 \right)\log 2+\dfrac{3x}{\left( x-1 \right)}\log 3=2\log 3 \\
\end{align}\]
Shifting all the terms to the L.H.S and simplifying by taking the common terms together we get,
\[\begin{align}
  & \Rightarrow \left( x+2 \right)\log 2+\dfrac{3x}{\left( x-1 \right)}\log 3-2\log 3=0 \\
 & \Rightarrow \left( x+2 \right)\log 2+\left( \dfrac{3x}{\left( x-1 \right)}-2 \right)\log 3=0 \\
 & \Rightarrow \left( x+2 \right)\log 2+\dfrac{\left( x+2 \right)}{\left( x-1 \right)}\log 3=0 \\
\end{align}\]
Taking (x + 2) common from both the terms we get,
\[\Rightarrow \left( x+2 \right)\left( \log 2+\dfrac{\log 3}{\left( x-1 \right)} \right)=0\]
Substituting each term equal to 0 one by one we get,
(i) When $\left( x+2 \right)=0$ we have,
$\therefore x=-2$
(ii) When \[\left( \log 2+\dfrac{\log 3}{\left( x-1 \right)} \right)=0\] we have,
\[\begin{align}
  & \Rightarrow \log 2=-\dfrac{\log 3}{\left( x-1 \right)} \\
 & \Rightarrow \left( x-1 \right)=-\dfrac{\log 3}{\log 2} \\
 & \therefore x=1-\dfrac{\log 3}{\log 2} \\
\end{align}\]

So, the correct answer is “Option a and b”.

Note: Note that you can use the base change formula to write $\dfrac{\log 3}{\log 2}={{\log }_{2}}3$ but in the options we have the former relation so we haven’t used the conversion. You can take log to any base because and not necessarily base 10 because at last we will get the fraction $\dfrac{\log 3}{\log 2}$ which can be changed into any base of log. Here both the values of x are valid so we need to consider both of them.