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.

2016年7月16日 · We can rewrite 9x2 = 27x by subtracting 27x from both sides and make it equal to 0. 9x2 = 27x 9x2 − 27x = 0 Factoring out a 9 and x we now get 9x(x − 3) = 0 We now have two separate factors, namely 9x = 0 and x − 3 = 0, so we can simply solve for x. 9x = 0 → x = 0 x − 3 = 0 → x = 3 So our solutions are x = 3 and x = 0.

2016年8月7日 · 3x2 − 27(2 − x)2 = 3x2 − 27(4 + x2 −4x) = 3x2 − 27(4) − 27x2 +27(4x) = 3x2 − 108 −27x2 + 108x = − 24x2 + 108x − 108 = − 12(2x2 −9x + 9) = − 12(2x2 −6x − 3x + 9) = − 12(2x(x −3) −3(x − 3)) = − 12(x −3)(2x −3)

2018年3月13日 · Raise to the power of 2 3 hence 273 2 = (x2 3)3 2 273 2 = x x = (271 2)3 x = (3√3)3 = 81√3

2015年7月22日 · 81 = 34 and 27 = 33, so this equation can also be written as 34x = 33(x+2) = 33x+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) are one-to-one functions).

2017年2月13日 · Using the transformation of 272 3 = (2 3) ⋅ log27 we calculate (2 3) ⋅ 1.431 = 0.9542 Taking the inverse (anti-log, or exponent) of this value gives us the answer: 100.9542 = 9

2017年6月11日 · change the base from 9 → 3 ⇒ (32)x = (33)x−2 ⇒ 32x = 33x−6 since the bases are the same, we can equate the exponents ⇒ 3x − 6 = 2x ⇒ x = 6

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.

2018年4月13日 · 3(3x-1)(3x+1) >"take out a "color(blue)"common factor "3 =3(9x^2-1) 9x^2-1" is a "color(blue)"difference of squares" •color(white)(x)a^2-b^2=(a-b)(a+b) "here "9x^2 ...

2015年8月19日 · 92x = 27x−1 We know that 9 = 32 27 = 33 So, 32(2x) = 33(x−1) 34x = 33x−3 Now as bases are equal we can equate exponents 4x = 3x − 3 x = − 3