Question

4^(4x-1)=32 How do I solve this problem? Do I do 4 to the fourth power first?

Answer 1

Answer:

$\large\boxed{x=\dfrac{7}{8}}$

Step-by-step explanation:

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

Answer 2

This is the answer to the problem