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.

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

2015年5月23日 · 3^(x-2)=27 27 can be written as 3^3 3^(x-2)=3^3 at this stage exponents can be equated x-2 = 3 color(blue)( x = 5 is the solution for the expression.

2015年7月22日 · More simply, x=6 (logarithms are unnecessary in this case). 81=3^{4} and 27=3^{3}, so this equation can also be written as 3^{4x}=3^{3(x+2)}=3^{3x+6}. Since the bases are now the same, we can equate exponents to get 4x=3x+6 so that x=6 (technically this equating of exponents requires the fact that exponential functions (with base not equal to 1 ...

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.#

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11月4日 · See a solution process below: First, solve for two points which solve the equation and plot these points: First Point: For x = 1 y = (-3/2 xx 1) - 27/2 y = -3/2 - 27/2 y = -30/2 y = -15 or (1, -15) Second Point: For x = -1 y = (-3/2 xx -1) - 27/2 y = 3/2 - 27/2 y = -24/2 y = -12 or (-1, -12) We can next plot the two points on the coordinate plane: graph{((x-1)^2+(y+15)^2 …

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

2017年7月22日 · Step 1) Subtract #color(red)(2)# and #color(blue)(x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

2015年5月23日 · x^2 -12x+27=0 We can Split the Middle Term of this expression to factorise it and thereby find the solution. In this technique, if we have to factorise an expression like ax^2 + bx + c, we need to think of 2 numbers such that: N_1*N_2 = a*c = 1*27 = 27 And N_1 +N_2 = b = -12 After trying out a few numbers we get N_1 = -9 and N_2 =-3 9*3 = 27 and -9+(-3)= -12 x^2 …