Question
Ive been staring at this problem for like a half hour now and I cant figure out where to even begin...
log(base x)4 + log(base 8)squareroot(x) = 4/3
solve for x
any help appreciated
log(base x)4 + log(base 8)squareroot(x) = 4/3
solve for x
any help appreciated
Answers
Reiny
Have you come across the property
log<sub>a</sub>b = log b/log a ?
log<sub>x</sub>4 + log<sub>8</sub>squareroot(x) = 4/3
log<sub>x</sub>4 + (1/2)log<sub>8</sub>(x) = 4/3
log4/logx + (1/2)logx/log8 = 4/3
let logx = t
log4/t + t/(2log8) = 4/3
multiply each term by (6log8)t to get the quadratic
6log4(log8) + 3t^2 = 8log8(t)
After re-arranging, use the quadratic formula to solve for t, after that sub back into logx = t to get x
check my steps, easy to make typos here.
log<sub>a</sub>b = log b/log a ?
log<sub>x</sub>4 + log<sub>8</sub>squareroot(x) = 4/3
log<sub>x</sub>4 + (1/2)log<sub>8</sub>(x) = 4/3
log4/logx + (1/2)logx/log8 = 4/3
let logx = t
log4/t + t/(2log8) = 4/3
multiply each term by (6log8)t to get the quadratic
6log4(log8) + 3t^2 = 8log8(t)
After re-arranging, use the quadratic formula to solve for t, after that sub back into logx = t to get x
check my steps, easy to make typos here.