126 = n^12
12 = log(base n)126
Since log(base n)(126) = log 126/log n or log(base n)(126) = ln 126/ln n we write:
12 = ln 126/ln n
12 ln n = ln 126
ln n = ln 126/12
ln n = 0.4030234922 rewrite the natural logarithm showing base e (optional)
log(base e)(n)= 0.4030234922
e^0.4030234922 = n
Check e^0.4030234922
126 = (e^0.4030234922)^12 ?
126 = e^4.836281907 ?
126 = 126 True
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Let the intgers be n- 1 n , & n+1 . This is consecutive. Hence Adding (n-1) + n + ( n +1) = 126 Add the LHS 3n = 126 Divide both sides by '3' n = 42 Hence n- 1 = 41 & n +1 = 43 So the consecutive integers are 41,42, & 43.
the answer would be exponentthe n in x indicating the number of factor of x is exponent
The exponential expression a^n is read a to the nth power. In this expression, a is the base and n is the exponent. The number represented by a^n is called the nth power of a.When n is a positive integer, you can interpret a^n as a^n = a x a x ... x a (n factors).
Any value with a 'zero' exponent is equaL TO '1'. A^(0) = 1 proof Let a^(0) =. a^(n - n) = a^(n) / a^(n) Cancel down by a^(n) hence it equals '1'.
Not necessarily. Every exponent in the exponent must be a non-negative integer. This is not what you have specified. For example, if n = 3.5, it is not a term in a polynomial expression.