Kenneth S. answered 06/17/16
Tutor
4.8
(62)
Let's cut to the chase: I know this subject & how to teach YOU
1) solve for x
a)log2√8 There's no x but we can evaluate; = log2(23/2) = 3/2
b)logx125=3/2 Put this log (exponent!) as exponent on base x;
a)log2√8 There's no x but we can evaluate; = log2(23/2) = 3/2
b)logx125=3/2 Put this log (exponent!) as exponent on base x;
125 = x(3/2); raise this to (2/3) power: (53)2/3 = x; x =52 = 25
c)log2(x-4) - log2(x+2)=4
c)log2(x-4) - log2(x+2)=4
log2[(x-4)/(x+2)] = 4;
exponentiate on base 2 ⇒(x-4)/(x+2)=16;
you can take it from here!
d)log2(x2-2x)7=21; 7log2[x(x-2)]=21;log2[x(x-2)]=3;
d)log2(x2-2x)7=21; 7log2[x(x-2)]=21;log2[x(x-2)]=3;
exponentiate on base 2 to get x(x-2)=9; this gives a quadratic equation that you can solve (be sure to check each answer).