2016年11月13日 · The answer is log_2(27) To solve this problem we need the inverse operation of exponents: logarithms. If 2 raised to some power x is equal to 27, then the logarithm of base 2 and an argument of 27 is equal to x.

2017年2月13日 · You can also use logarithmic functions, if you have a logarithm table or at least a calculator that can do log functions, if not complex exponential ones.

2015年7月8日 · color(blue)(x=2 9^(2x-1) = 3^(2.(2x-1)) = 3^(4x-2) 27^x = 3^(3x) Upon equating: 3^(4x-2) = 3^(3x) Since the base is equal , we can equate the powers: 4x-2 =3x color ...

2017年7月4日 · You may start by dividing everything by 3 ->x^2-8x+9=0 You take half of the x coefficient and square that. The square would then be: x^2-8x+(-4)^2=x^2-8x+16=(x-4)^2 To get there you would have added 7 to the original 9.

2015年2月1日 · x²−27= (x)²- (3√3)²=(x+3√3)(x-3√3)

2017年6月11日 · x=6 "change the "color(blue)"base "" from " 9 to 3 rArr(3^2)^x=(3^3)^(x-2) rArr3^(2x)=3^(3x-6) "since the bases are the same, we can equate the exponents" rArr3x-6=2x rArrx=6

2016年6月29日 · Given Eqn. : #x^(3/2)=27.# #:.{x^(3/2)}^2=27^2.# #:. x^{3/2*2}=(3^3)^2.# #:.x^3=3^(3*2)=3^6=(3^2)^3.# As powers are same, so must be the bases, so, #x=3^2=9.# It can be easily verified that the root found satisfies the given eqn. Hence, the Soln. is #x=9.#

2018年2月19日 · \qquad "the points of inflection are:" \qquad (-3, 6), \quad (3, 6). # "First, we need to find" \ \ f' '(x) "."

2015年8月20日 · #x^2−6x−27=0# We can Split the Middle Term of this expression to factorise it and thereby find solutions.. In this technique, if we have to factorise an expression like #ax^2 + bx + c#, we need to think of 2 numbers such that:

2016年11月3日 · Because factoring x^3+27 is the same as finding where the graph passes through the x axis, we can just set the equation equal to zero and solve. f(x) = (x+3)(x^2-3x+9) Let f(x) = x³ + 27 0 = x³ + 27 x³= -27 x = -3 This means that x = -3 is the only zero of the graph of f(x). Since we know (x+3) is one factor of f(x), to find the 2nd factor, (x^3+27)/(x+3) We get the second factor to be x^2 ...