Chapter 6.4, Problem 17E

To determine

To solve the equation:

  ${36}^{-3x+3}={\left(\frac{1}{216}\right)}^{x+1}$

And check the solution.

Answer to Problem 17E

  $x=3$

Explanation of Solution

Given:

  ${36}^{-3x+3}={\left(\frac{1}{216}\right)}^{x+1}$

Concept Used:

• $x^{-n}=\frac{1}{x^n};x\ne 0$
• Two powers with the same positive base(not equals to 1), are equal if and only if their exponents are equal, that is:

  $b^x=b^y\iff x=y;\text{ where }b>0,\text{ and }b\ne 1$

Calculation:

To solve the equation:

  ${36}^{-3x+3}={\left(\frac{1}{216}\right)}^{x+1}$

First, simplify the equation so the base of the equation is same, as

  $\begin{array}{c}{\left(6^2\right)}^{-3x+3}={\left(\frac{1}{6\times 6\times 6}\right)}^{x+1}\\ \Rightarrow 6^{2\left(-3x+3\right)}={\left(\frac{1}{6^3}\right)}^{x+1}\\ \Rightarrow 6^{-6x+6}={\left(6^{-3}\right)}^{x+1}\\ \Rightarrow 6^{-6x+6}=6^{-3\left(x+1\right)}\\ \Rightarrow 6^{-6x+6}=6^{-3x-3}\end{array}$

Now, since base of above powers are same; thus

  $\begin{array}{c}{6}^{-6x+6}=6^{-3x-3}\\ \Rightarrow -6x+6=-3x-3\\ \Rightarrow 6+3=-3x+6x\\ \Rightarrow 9=3x\\ \Rightarrow 3x=9\\ \Rightarrow x=\frac{9}{3}\\ \Rightarrow x=3\end{array}$

Thus, the solution of given equation is:

  $x=3$

Now, to check this solution, first substitute $x=3$ in left side of the given equation, as

  $\begin{array}{c}{36}^{-3x+3}={36}^{-3\left(3\right)+3}\\ ={36}^{-9+3}\\ ={36}^{-6}\\ ={\left(6^2\right)}^{-6}\\ =6^{2\times \left(-6\right)}\\ =6^{-12}\end{array}$

Now, substitute $x=3$ in right side of the given equation, as

  $\begin{array}{c}{\left(\frac{1}{216}\right)}^{x+1}={\left(\frac{1}{216}\right)}^{3+1}\\ ={\left(\frac{1}{6^3}\right)}^4\\ ={\left(6^{-3}\right)}^4\\ =6^{-12}\\ =\text{Left side}\end{array}$

Since, $x=3$ gives both sides of the equation equal, i.e. $x=3$ is correct solution of the given equation.