Problem: The real value of xxx such that 64xβ164^{x-1}64xβ1 divided by 4xβ14^{x-1}4xβ1 equals 2562x256^{2 x}2562x is:
Answer Choices:
A. β23-\dfrac{2}{3}β32β
B. β13-\dfrac{1}{3}β31β
C. 000
D. 14\dfrac{1}{4}41β
E. 38\dfrac{3}{8}83β Solution:
64xβ1Γ·4xβ1=43xβ3Γ·4xβ1=42xβ264^{x-1} \div 4^{x-1}=4^{3 x-3} \div 4^{x-1}=4^{2 x-2}64xβ1Γ·4xβ1=43xβ3Γ·4xβ1=42xβ2. Since 2562x=48x,42xβ2=48xβ΄2xβ2=8x,x=β13256^{2 x}=4^{8 x}, 4^{2 x-2}=4^{8 x} \quad \therefore 2 x-2=8 x, x=-\dfrac{1}{3}2562x=48x,42xβ2=48xβ΄2xβ2=8x,x=β31β.