Question

How do you solve the exponential equation \[{{2}^{x+2}}={{16}^{x}}\]?

Answer 1

Hint: Write 16 in the R.H.S. in exponential form with base equal to 2. Now, apply the formula, \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] to simplify the R.H.S. Now, compare the bases on both the sides and equate the exponents to form a linear equation in x. Solve this equation for the value of x to get the answer.

Complete step by step answer:
Here, we have been provided with the exponential expression: - \[{{2}^{x+2}}={{16}^{x}}\] and we are asked to solve it. That means we have to find the value of x.
Now, we can write 16 in the R.H.S. as \[16={{2}^{4}}\], so we have,
\[\Rightarrow {{2}^{x+2}}={{\left( {{2}^{4}} \right)}^{x}}\]
Using the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\Rightarrow {{2}^{x+2}}={{2}^{4x}}\]
As we can see that both the sides of the above exponential expression contains 2 as the bases, so we can compare and equate the exponents by removing the bases from both the sides. So, we have,
\[\begin{align}
  & \Rightarrow x+2=4x \\
 & \Rightarrow 4x-x=2 \\
 & \Rightarrow 3x=2 \\
\end{align}\]
Dividing both the sides with 3, we get,
\[\Rightarrow x=\dfrac{2}{3}\]

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 2, i.e., \[{{\log }_{2}}\] on both the sides and use the property \[{{\log }_{n}}n=1\] to get the answer. Here, n > 0 and \[n\ne 1\].