Question 634369
solution for 2logg(4-x) to the base 4=4-log(-2-x) to the base 2
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2log4(4-x)=4-log2(-2-x)
2log4(4-x)+log2(-2-x)=4
convert to log base2
2log2(4-x)/log2(4)+log2(-2-x)=4
log2(4-x)+log2(-2-x)=4
log2[(4-x)(-2-x)]=4
convert to exponential form
2^4=(4-x)(-2-x)
16=-8-2x+x^2
x^2-2x-8-16=0
x^2-2x-24=0
(x-6)(x+4)=0
x=6 (reject,(-2-x)>0
or
x=-4
check:
2log4(4-x)=4-log2(-2-x)
2log4(4+4)=4-log2(2)
log4(8^2)=4-1
log4(64)=3
3=3