Question 28485
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