Chapter 6.5, Problem 46E

To determine

To solve the equation:

  $5^{8\left(x-1\right)}={625}^{2x-2}$

Answer to Problem 46E

Infinitely many solutions

Explanation of Solution

Given:

  $5^{8\left(x-1\right)}={625}^{2x-2}$

Concept Used:

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:

  $5^{8\left(x-1\right)}={625}^{2x-2}$

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

  $\begin{array}{c}{5}^{8\left(x-1\right)}={\left(5\times 5\times 5\times 5\right)}^{2x-2}\\ \Rightarrow 5^{8\left(x-1\right)}={\left(5^4\right)}^{2x-2}\\ \Rightarrow 5^{8x-8}=5^{4\left(2x-2\right)}\\ \Rightarrow 5^{8x-8}=5^{8x-8}\end{array}$

Now, since base of above powers are same; thus

  $\begin{array}{c}{5}^{8x-8}=5^{8x-8}\\ \Rightarrow 8x-8=8x-8\\ \Rightarrow 8x-8x=8-8\\ \Rightarrow 0=0\end{array}$

Now, the statement $0=0$ is true for every value of variable x . Thus, the given equation is an identity.

Thus, the given equation has infinitely many solutions.