Question

Pls answer this...its very important for me...i have an exam coming up...and this 2 questions will come up so pls help. I WILL MARK AS BRAINLIEST WHOEVER ANSWERED RIGHT.
1. Find the value of x
a) $\frac{4^{2x} }{4^{x} } = 64$
b) $(2^{2x} )^{3} = 64$

Answer 1

Answer:

a. x = 3

b. x = 3

Step-by-step explanation:

Given

$\frac{4^{2x}}{4^x} = 64$

$(2^{2x}) = 64$

Required

Find x

Solving $\frac{4^{2x}}{4^x} = 64$

From laws of indices;

$\frac{a^{m}}{a^n} = a^{m-n}$

So;

$4^{2x-x} = 64$

$4^{x} = 64$

Write 64 in base of 4

$4^{x} = 4^3$

From laws of indices;

$if\ x^a = x^b;\ then\ a = b$

So; x = 3

Solving $(2^{2x}) = 64$

Write 64 as a base of 2

$(2^{2x}) = 2^6$

$if\ x^a = x^b;\ then\ a = b$

$2x = 6$

Divide both sides by 2

$\frac{2x}{2} = \frac{6}{2}$

$x = 3$