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 …

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 …

2016年7月16日 · x = 3 and x = 0 We can rewrite 9x^2 = 27x by subtracting 27x from both sides and make it equal to 0. 9x^2 = 27x 9x^2 - 27x = 0 Factoring out a 9 and x we now get 9x(x-3) …

2018年3月13日 · Raise to the power of 2/3 hence 27^(3/2)=(x^(2/3))^(3/2) 27^(3/2)=x x = (27^(1/2) )^3 x = (3sqrt(3))^3 = 81sqrt(3)

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.

2018年4月13日 · Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one …

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 …

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

2015年8月19日 · color(blue)(x=-3 9^(2x)=27^(x−1 We know that 9=3^2 27=3^3 So, 3^(2(2x))=3^(3(x−1) 3^(4x)=3^(3x−3 ...