Chapter 8.2, Problem 48HP

To determine

To proof: The expression ${27}^{2x}\cdot {81}^{x+1}=3^{2x+2}\cdot 9^{4x+1}$ holds true for every x.

Explanation of Solution

Given information:

The expression ${27}^{2x}\cdot {81}^{x+1}=3^{2x+2}\cdot 9^{4x+1}$ .

Formula used:

Summation property:

If $b^x\cdot b^y=b^u\cdot b^v$ then $b^{x+y}=b^{u+v}$for all $b>0$ ,and $b\ne 1$ .

Proof:

Consider the given expression ${27}^{2x}\cdot {81}^{x+1}=3^{2x+2}\cdot 9^{4x+1}$ .

Since 27 can be written as $3^3$ , also the number 81 and 9 can be written as $3^4$ and $3^2$ respectively. So, rewrite the given equation as like this,

  $\begin{array}{c}{27}^{2x}\cdot {81}^{x+1}=3^{2x+2}\cdot 9^{4x+1}\\ 3^{3\left(2x\right)}\cdot 3^{4\left(x+1\right)}=3^{2x+2}\cdot 3^2{}^{\left(4x+1\right)}\\ 3^{6x}\cdot 3^{4x+4}=3^{2x+2}\cdot 3^{8x+2}\end{array}$

Apply the summation property,

  $\begin{array}{c}{3}^{6x}\cdot 3^{4x+4}=3^{2x+2}\cdot 3^{8x+2}\\ 3^{6x+4x+4}=3^{2x+2+8x+2}\\ 3^{10x+4}=3^{10x+4}\end{array}$

Since left hand side is equal to right hand side.

Hence, the expression ${27}^{2x}\cdot {81}^{x+1}=3^{2x+2}\cdot 9^{4x+1}$ holds true for every x.