Question

4^(4x-1)=32

How do I solve this problem? Do I do 4 to the fourth power first?

Answer 1

This is the answer to the problem

Answer 2

`\large\boxed{x=(7)/(8)}`

Explanation:

`4^((4x-1))=32\\\\(2^2)^(4x-1)=2^5\qquad\text{use}\ (a^n)^m=a^(nm)\\\\2^(2(4x-1))=2^5\iff2(4x-1)=5\ \ \text{use the distributive property}\ a(b+c)=ab+ac\\\\(2)(4x)+(2)(-1)=5\\\\8x-2=5\qquad\text{add 2 to both sides}\\\\8x=7\qquad\text{divide both sides by 8}\\\\x=(7)/(8)`