Find the value of x in: 3 log (base x) 2 - 1/2 log (base x) 16 = 5?

log x is denoted by " log " in the following:-
3 log 2 - (1/2) log 16 = 5
log 2 ³ - log 16^(1/2) = 5
log 8 - log 4 = 5
log (8/4) = 5
log 2 = 5
logx 2 = 5
2 = x^5
x = 2^(1/5)

5 = log (base x) x^5

2^3/sqrt(16) = x^5

8/ 4 = x^5

x= ...

2. 0, assuming the domain of x is restricted to positive integers. Otherwise, (-3/5)^x increases without bounds for x is even and negative. Any non-integral value of x would give an undefined result; any odd power would give a negative result, any even power greater than 0 would give a result between 0 and 1, but any number (besides 0) raised to the zero power is 1. 3. 18 5^15/5^n = 1/125 5^(15 - n) = 1/5^3 5^(15 - n) = 5^(-3) 15 - n = -3 n = 18 5. 4 log5 (32) - 4log5 (m) = 3 log5 (1/2) log5 (32) - log5 (m^4) = log5 (1/2^3) log5 (32/m^4) = log5 (2^-3) 32/m^4 = 2^-3 32 = 2^-3 * m^4 m^4 = 32 * 2^3 m^4 = 2^5 * 2^3 = 2^(5 + 3) = 2^8 m^4 = 2^(2 * 4) m^4 = (2²)^4 = 4^4 m = 4 6. 4 (3 + 7i)(5 - yi) = (3)(5) + (3)(-yi) + (7i)(5) + (7i)(-yi) = 15 - 3yi + 35i - 7yi² i² by definition is -1, so we have 15 - 3yi + 35i - 7y(-1) = (15 + 7y) + (-3y + 35)i For this to equal 43 + 23i and have y be a real number, both the real part and the imaginary part must be equal. Set the real part equal: 15 + 7y = 43 7y = 43 - 15 = 28 y = 28/7 = 4 Check this with the imaginary part: If y = 4, then (-3y + 35)i = (-3(4) + 35)i = (-12 + 35)i = 23i, so y = 4 is correct. 6. 2.5 log5 (25√5) = log5 (5 * 5 * 5^(1/2)) = log5 (5² * 5^(1/2)) = log5 (5^(2 + 1/2)) = log5 (5^2.5) = 2.5 * log5 (5) = 2.5 * 1 = 2.5

x^5 = 2^3/root(16) = 2

x = 2^(1/5)

c'est tout !!!

3 log (base x) 2 - 1/2 log (base x) 16 = 5
Or log(base x){2^3 / 16^(1/2)} = 5
Or {2^3 / 16^(1/2)} = x^5
Or x^5 = 2
Or x = 2^(1/5)