Posted by anonymous on Wednesday, January 9, 2013 at 12:05pm.
solve the equation
log2(x+4)log4x=2
the 2 and 4 are lower than the g
This is what I got:
log2(x+4)+log2(4^x)=2
log2((x+4)*4^x)=2
4^x(x+4)=4
x=0 is a solution???

math  justastudent, Wednesday, January 9, 2013 at 12:51pm
log2(x+4) + log4(x) = 2
log2(x+4) + log2(x)/log2(4) = 2
log2(x+4) + log2(x)/2 = 2
2 log2(x+4) + log2(x) = 4
log2((x+4)^2) + log2(x) = 4
log2(x(x+4)^2) = 4
x (x+4)^2 = 2^4
x^3 + 8x^2 + 16x − 16 = 0
x = 0.718608