2016年1月18日 · x=4 or x=-4 First divide throughout by 2 to get x^2=16 Now take the square root on both sides to get x=+-4. You may now substitute the values of x=4 and x=-4 back into the original equation and check that it satisfies the equation which it does.

2015年10月26日 · x= 5 or -1 Looking at your format I assume you meant to write: 2^ (x^2) = 32 (2^(4x)) => (2^ (x^2))/(2^(4x)) = 32 => 2^((x^2 - 4x)) = 32 Taking logs: =>(x^2 - 4x)log ...

2018年3月22日 · x=4sqrt(2)" " Exact answer for principle root x=+-4sqrt(2)" " As a full exact answer x+-5.6569 to 4 decimal places - Approximate answer Consider the 32. We are looking for squared number we can 'take outside' the root 32 is even so we have 2xx16 But 16 is 4^2 giving 2 xx 4^2. So we can write: x=sqrt(2xx4^2) x=4sqrt(2) Using the Principle Root which is always …

2016年1月10日 · -32/x^2 Before differentiating this function it requires to be written in the form ax^n note that the derivative of a function must be in the correct form before you can differentiate it. f(x) = ax^n rArr f'(x) =nax^(n - 1 ) 32/x = 32x^-1 now differentiating gives -1xx32x^-2 rewriting with a positive index the gives -32/x^2

2016年3月9日 · x=5 The logarithmic expression can be exponentiated with a base of 2: 2^(log_2 32)=2^x The 2^x and log_2 x functions are inverses, which means that they undo one another, so we obtain the equation: 32=2^x We can write 32 as a power of 2: 2^5=2^x Since the bases are equal, we know their powers are also equal, giving: x=5

2016年3月21日 · color(blue)((x + 2) ( x -16) is the factorised form of the expression. y = x^2 - 14x - 32 We can Split the Middle Term of this expression to factorise it. 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 * (-32) = -32 AND N_1 +N_2 = b = -14 After trying out a few numbers we get N_1 = -16 …

2015年11月23日 · color(blue)((x-4)(x-8) is the factorised form of the expression. x^2-12x+32 We can Split the Middle Term of this expression to factorise it. 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*32 =32 AND N_1 +N_2 = b = -12 After trying out a few numbers we get: N_1 = -8 and N_2 =-4 ( …

2016年5月20日 · -32/x^2 >Begin by expressing 32/x=32x^-1 differentiate using the color(blue)"power rule" d/dx(ax^n)=nax^(n-1) rArrd/dx(32x^-1)=-1xx32x^-2=-32/x^2

2017年4月29日 · x=3 First of all, you should know that any number of the form a^(bc) can be written in the form (a^b)^c or (a^c)^b or vice-versa. i.e. a^(bc) = (a^b)^c = (a^c)^b. Also, a^(b+c) = a^b*a^c therefore above equation 2^(2x)-(2^(x+2))-32=0 can be writte as:- (2^x)^2 - 2^2*2^x-32=0 => (2^x)^2-4*2^x-32=0 Let 2^x be some number y. therefore y^2-4y-32=0 Now solving it just …

2016年3月8日 · Vertical asymptote: x=0 Horizontal asymptote: y=0 To find vertical asymptotes, set the denominator of the function equal to 0. x^2=0 x=0 There is a vertical asymptote at x=0. The rule for finding horizontal asymptotes when the degree of the denominator is larger than the numerator, as is the case here, since 2>1, the horizontal asymptote will be the line y=0. …