Question

If we are given as expression as \[{{3}^{x}}\times {{27}^{x}}={{9}^{x+4}}\], then what is x equal to?
(a) 4
(b) 5
(c) 6
(d) 7

Answer 1

Hint: Write 27 in the L.H.S and 9 in the R.H.S as exponents of 3. In the L.H.S apply the formula \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] to convert the expression containing a single base 3. In the R.H.S apply the formula \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] to simplify the expression. Now, compare the bases on both sides and equate the exponents to form a linear equation in x. Solve the equation to get the value of x.

Complete step-by-step solution
Here, we have been provided with the expression: - \[{{3}^{x}}\times {{27}^{x}}={{9}^{x+4}}\]. So, we can write,
\[\Rightarrow 27={{3}^{3}}\] and \[9={{3}^{2}}\]
Therefore, the above expression can be written as,
\[\Rightarrow {{3}^{x}}\times {{\left( {{3}^{3}} \right)}^{x}}={{\left( {{3}^{2}} \right)}^{x+4}}\]
Applying the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
  & \Rightarrow {{3}^{x}}\times {{3}^{3x}}={{3}^{2\left( x+4 \right)}} \\
 & \Rightarrow {{3}^{x}}\times {{3}^{3x}}={{3}^{2x+8}} \\
\end{align}\]
Now, applying the formula: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], we get,
\[\begin{align}
  & \Rightarrow {{3}^{x+3x}}={{3}^{2x+8}} \\
 & \Rightarrow {{3}^{4x}}={{3}^{2x+8}} \\
\end{align}\]
Here, we can clearly see that in the above relation both sides contain 3 as the base of the two terms. So, for the two terms to be equal their exponents must be equal. Therefore, on comparing their exponents, we get,
\[\begin{align}
  & \Rightarrow 4x=2x+8 \\
 & \Rightarrow 4x-2x=8 \\
 & \Rightarrow 2x=8 \\
 & \Rightarrow x=4 \\
\end{align}\]
Hence, option (a) is the correct answer.

Note: One may note that here we have used some basic formulas of the topic ‘exponents and powers’ to solve the question. You must remember some basic formulas such as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] and \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] because they are used everywhere. We can also solve the above question with the help of logarithm. We can take log to the base 3, i.e. \[{{\log }_{3}}\], both sides and use their properties to get the answer.