How would I do this?..

1. 3 logb(2) x=12

2. 3+4 logb(x) 4=5

1. 3 log(base 2) x =12
log (base 2) x^3 = 12
Taking anti-log
x^3 = 2^12 = (2^4)^3
taking cube root
x =2^4=16
Hence x =16 ................Ans

2. 3+4(log base x) 4 =5
4 (log( base x) 4 = 5-3 =2
log(base x) 4^4= 2
taking anti-log
4^4 = x^2
x^2 = 256
x = 16 ..................Ans

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➊   3 log₂x = 12            ← first, divide both sides by 3
        log₂x = 4
              x = 2⁴
              x = 16


➋   3 + 4 logₓ4 = 5            ← first, subtract 3 from both sides
           4 logₓ4 = 2
          logₓ(4⁴) = 2
          logₓ(4⁴) = 2
                   x² = 4⁴            ← now, take the square root of both sides
                    x =  ±√(4⁴)            ← by the square root property
                    x = ±4²
                    x = ±16        ← Notice that the base of the log in the original equation is x.
                                            And, since log bases must be positive, we discard  -16.
                  ANSWER
                     x = 16


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