MarthaAndrews MarthaAndrews
  • 04-07-2019
  • Mathematics
contestada

4^(4x-1)=32

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

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gmany
gmany gmany
  • 04-07-2019

Answer:

[tex]\large\boxed{x=\dfrac{7}{8}}[/tex]

Step-by-step explanation:

[tex]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}[/tex]

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crystalbellofosu
crystalbellofosu crystalbellofosu
  • 04-07-2019
This is the answer to the problem
Ver imagen crystalbellofosu
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