Question 28485: solve these
2log4x-log16=1
3lnx+lnx=4
3(x)=243
e(x+4)=48
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! solve these
2log4x-log16=1
3lnx+lnx=4
3(x)=243
e(x+4)=48
)2log4x-log16=1
log[(4x)^2]-log(16) = 1 [using nlog(m) = log(m^n)]
log(16x^2)-log(16) = 1
log[(16x^2)/(16)] = 1 [using loga - log b = log(a/b) ]
log(x^2)=1
(x^2) = (10)^1 [using logb(N) = p implying N = (b)^p ]
x^2 = 10
x=+[sqrt(10)]
Negative sqrt cannot hold here because when you use the negative value for x, you get into a situation where you log(of a negative quantity) which is not defined.
2)3lnx+lnx=4
log[(x)^3]+log(x)= 4 [using nlog(m) = log(m^n)]
log[(x^3)X(x)]=4 [using loga + log b = log(ab)]
log(x^4)=4
x^4 = 10^4
x=10 (powers equal imply bases equal)
3(x)=243 this simply implies x = 243/3 = (3X81)/3 = 81
If the problem is 3log(x) =243
which means log(x) = 243/3 = 81
logx = 81
gives x = (10)^81 (quite a BIG number!)
4)e(x+4)=48
(x+4) = 48/e
x = (48/e) -4
This problem too may not be the correct original problem.
Please check
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