Negative Exponential Functions?
Let's simplify your question to f(x)=-b^x where b is a positive number and only consider positive, integral values of x. f(0) = 0 f(1) = -b f(2) = b^2 f(3) = -b^3 f(4) = b^4 f(5) = -b^5 As you see, this oscillates between positive and negative numbers constantly, meaning that the function is not continuous. Furthermore, now let's consider non-integer values of x. f(3.3) = -b^3.3 but..... is b^3.3 positive or negative? It's undefined. This is why b must be positive
Math, exponential function with a negative exponent and variable?
31 = 89.976 * e^(- 0.023x) 89.976 * e^(- 0.023x) = 31 e^(- 0.023x) = 31/89.976 Ln[e^(- 0.023x)] = Ln(31/89.976) → recall: Ln(x^a) = a.Ln(x) - 0.023x.Ln(e) = Ln(31/89.976) → recall: Ln(e) = 1 - 0.023x = Ln(31/89.976) x = - Ln(31/89.976) / 0.023 → recall: Ln(a/b) = - Ln(b/a) x = Ln(89.976/31) / 0.023 x ≈ 46.32851146
Integration of an exponential function with negative exponent?
∫e^(x).dx = e^(x). ∫e^(ax).dx = 1/a[e^(ax)]. . . . . . . this is the 'rule' that you need to know that is 'close'! So, if a = -1: ∫e^(-x).dx = -e^(-x). {I have left out the 'constant of integration' for simplicity} .
What do the variables mean in the exponential function?
In the exponential function f(x) = a * b^(cx) + d x: the input or the independent variable a: Amplitude multiplier, it scales the exponential function. b: Basis for the exponential function c: Multiplier for x, both cx is the exponent of the exponential function. They control the rate of change of the exponent, that is how fast the exponential function is changing with respect to x. If it is negative, it controls 1/2 life, if it is positive it controls how fast the exponent double...etc. d: the offset In general, the graph of f(x) = a*b^(cx) + d has a y-intercept at (0,a+d) and is asymptotic to the x-axis from above if a is positive and from below if a is negative. If b is greater than 1, then the graph is asymptotic to the x-axis as x decreases without bound and increases without bound as x increases without bound. If b is less than 1, then the graph is asymptotic to the x-axis as x increases without bound and increases without bound as x decreases without bound.
How do you integrate a variable with a negative exponent?
Just like with a positive exponent, add 1 to the power and divide by the new power. In your example add 1 to -3 and divide by -2 to get: -0.5x^-2 + 5x + C ------------------------------------ the only time you cannot do this is when the power is -1 which is a special result: ax^-1 = a*ln|x|
Solve for a variable with a negative exponent?
well.. (y/50)^(-1/2)=q^2+1 so q=+or- root[(y/50)^(-1/2)-1] ps. always take conjugates to get rid of unwanted exponents:)
How do you solve an exponential equation with a variable in the base?
How do you solve an exponential equation with a variable in the base?You mean something like [math]x^a=b[/math] with [math]a[/math] and [math]b[/math] given numbers? Simply take the [math]a[/math]-root from [math]b[/math]. [math]x=\sqrt[a]{b}=b^{\frac{1}{a}}[/math].More general “exponential equation with a variable in the base” are called polynomials and you can solve them up to the 4th degree (4 is the highest exponent and only natural exponents), now solving a polynomaial of the first degree (linear equation) and second degree (quadric equation) is easy, for 3 and 4 degree you can surely find methods by looking.
Math Help, Exponential Functions?
Your variable is stuck in your exponent, so you need to "throw a log" at both sides (as my alg 2 teacher puts it) and attach a logarithm base 10 to each side.. After you do that you can rearrange it so it looks like this: (2x^2+5x-3) log 3 = log 1 Then if you divide each side by log 3 you get: (2x^2)+(5x)-3 = log 1/log 3 Since log 1 is 0, and zero divided by anything is zero, you get: 2(x^2)+5x-3= 0 Now you can use the quadratic formula: [-b +- sqrt(b^2 - 4ac)]/2a (and when I say +- I don't mean plus a negative I mean plus OR minus) a=2, b=5 c=-3 [-5+- sqrt(25+24)]/4 [-5+- sqrt(49)]/4 (-5+-7}/4 (-5+7)/4, (-5-7)/4 Your roots (the places where your graph will cross the x axis) are 1/2 and -3.
What is the parent function for exponential functions?
Exponential functions stem from the mother function:[math]p(x)=(b)^x[/math] For [math]b>0[/math] and [math]b\neq 1[/math]The most general function would be:[math]g(x)=a(b)^{c(x-h)}+k[/math]Where a, would affect how fast the function approaches the horizontal asymptote for the base b, when a is negative then we have a vertical flip about the horizontal asymptote.b, is the base, so is first number that will affect how fast the graph will approach the horizontal asymptote. Bigger values of b will make the exponential approach infinity faster, while numbers between 0 and 1 in this place will have your exponential decay towards the asymptote (from left to right), and values closer to 0 will decay faster.c, will directly affect how wide the graph is for any given window, if c is negative then the exponential will have a horizontal flip about the y-axish, will “slide” or move the graph of the exponential to the left or right, normally when the exponent is zero, the graph will be 1 unit above the horizontal asymptote, so by changing h, you can change WHEN the exponent is zero, and thus shift left and right. (it’s nice to note that if c=1 and exponent is x-5, then it is a right shift by 5, not left as one might intuit.k, will shift the graph up and down, k is easily thought of as the placement for the horizontal asymptote, which is at y=k.