# SOLUTION: solve these 2log4x-log16=1 3lnx+lnx=4 3(x)=243 e(x+4)=48

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 Click here to see ALL problems on Exponential-and-logarithmic-functions Question 28485: solve these 2log4x-log16=1 3lnx+lnx=4 3(x)=243 e(x+4)=48Answer 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