Question

How do you solve \[{{3}^{x-2}}=81\]?

Answer 1

Hint: This question belongs to the topic of pre-calculus. In solving this question, first we will do the prime factorization of 81. After the factorization, we will balance the equation in such a way that the bases of both sides of the equation are equal. After that, we will use the formula \[{{x}^{a}}\times {{x}^{b}}={{x}^{\left( a+b \right)}}\] and solve the further equation and get the value of x. After that, we will see the alternate method to solve the question.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to solve the term \[{{3}^{x-2}}=81\]. We have to find out the value of x using the given equation \[{{3}^{x-2}}=81\].
The equation we have to solve is
\[{{3}^{x-2}}=81\]
Let us factorize the term 81.
\[\begin{align}
  & 3\left| \!{\underline {\,
  81 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
From the above factorization, we can write
\[81=3\times 3\times 3\times 3\]
The equation \[{{3}^{x-2}}=81\] can also be written as
\[{{3}^{x-2}}=3\times 3\times 3\times 3\]
\[\Rightarrow {{3}^{x-2}}={{3}^{1}}\times {{3}^{1}}\times {{3}^{1}}\times {{3}^{1}}\]
Using the formula \[{{x}^{a}}\times {{x}^{b}}={{x}^{\left( a+b \right)}}\], we can write the above equation as
\[\Rightarrow {{3}^{x-2}}={{3}^{\left( 1+1+1+1 \right)}}\]
\[\Rightarrow {{3}^{x-2}}={{3}^{4}}\]
Here, we can see that the bases are equal on both sides, then we can say that power or indices will also be equal. So, we can write
\[\Rightarrow x-2=4\]
The above equation can also be written as
\[\Rightarrow x=4+2\]
\[\Rightarrow x=6\]
Now, we have solved the equation \[{{3}^{x-2}}=81\] and have found the value of x as 6.

Note:
We should have a better knowledge in the topic of pre-calculus to solve this type of question easily. Always remember the following formula:
\[{{x}^{a}}\times {{x}^{b}}={{x}^{\left( a+b \right)}}\]
And, also remember that if bases are same that is in mathematical form we can write \[{{x}^{a}}={{x}^{b}}\], then their powers or indices will be same that is \[a=b\].
We should know how to factorize a number.
We have an alternate method to solve this question.
The equation is
\[{{3}^{x-2}}=81\]
Taking \[{{\log }_{e}}\] (\[\log \] to the base e) or we can say \[\ln \] to the both side of equation, we get
\[\Rightarrow \ln \left( {{3}^{x-2}} \right)=\ln \left( 81 \right)\]
As we know that 81 can be written as 3 to the power of 4, so we can write
\[\Rightarrow \ln \left( {{3}^{x-2}} \right)=\ln \left( {{3}^{4}} \right)\]
Using the formula \[\ln {{a}^{x}}=x\ln a\], we can write the above equation as
\[\Rightarrow \left( x-2 \right)\ln \left( 3 \right)=4\ln \left( 3 \right)\]
\[\Rightarrow \left( x-2 \right)\ln 3=4\ln 3\]
Now, the term \[\ln 3\] will be cancelled out to the both side of equation, we get
\[\Rightarrow \left( x-2 \right)=4\]
\[\Rightarrow x-2=4\]
The above equation can also be written as
\[\Rightarrow x=4+2=6\]
Hence, we get the value of x as 6 from here too. So, we can use this method too.