1.find the zeroes of the function f(x)=log4(x+1)+log4(4x-3)
2.which of the following is the equation c^(4d+1)=7a-b written in logarithm form
log4(x+1)+log4(4x-3) = 0
log4 (x+1)(4x-3) = 0
since 4^0 = 1, we have
(x+1)(4x-3) = 1
4x^2+x-3 = 1
4x^2+x-4 = 0
x = (-1±√65)/8
check for extraneous roots, since the domain of logz is z>0
c^(4d+1)=7a-b
logc(7a-b) = 4d+1
steve is right! thank you steve!
1. Oh, finding the zeroes, huh? Well, this function reminds me of a circus act! So, let's solve it, tightrope style! We have f(x) = log₄(x+1) + log₄(4x-3). First, let's get rid of those logs by using some circus magic! We can rewrite it as 4^(f(x)) = (x+1)(4x-3). Now, let's find the zeroes by making the left side zero. Oh, I guess that makes us tightrope walkers AND magicians now! Anyway, we set 4^(f(x)) = 0, but wait, a clown just popped out saying, "Hey, you can't raise 4 to the power of anything and get zero!" Silly clown, always showing up at the wrong time. So, long story short, there are no zeroes for this function!
2. Ah, another question! Alright, let's take this mathematical circus act further! The equation c^(4d+1) = 7a - b is quite the magician, switching between exponential and logarithmic forms. To reveal its true form, let's wave our magic wand and transform it! Here we go! In logarithmic form, it would be logₚ(7a - b) = 4d + 1. Abracadabra! Now, it's standing on the stage, ready for applause and little clown noses.
1. To find the zeros of the function f(x) = log4(x+1) + log4(4x-3), we need to solve the equation f(x) = 0.
First, let's simplify the equation by using the properties of logarithms. The sum of two logarithms with the same base can be written as the logarithm of their product. So, we can rewrite the equation as log4((x+1)(4x-3)) = 0.
Now, since log4(1) = 0 for any base, we can set the argument of the logarithm equal to 1 and solve for x. So, we have (x+1)(4x-3) = 1.
Expanding the equation, we get 4x^2 + x - 3 = 1.
Rearrange the equation and combine like terms: 4x^2 + x - 4 = 0.
Next, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, factoring is not straightforward, so we will use the quadratic formula.
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = 4, b = 1, and c = -4. Plugging these values into the quadratic formula, we get x = (-1 ± √(1^2 - 4*4*(-4))) / (2*4).
Simplifying further, x = (-1 ± √(1 + 64)) / 8.
x = (-1 ± √65) / 8.
Therefore, the zeroes of the function f(x) = log4(x+1) + log4(4x-3) are (-1 + √65) / 8 and (-1 - √65) / 8.
2. To write the equation c^(4d+1) = 7a - b in logarithm form, we can express it as log(base c) (7a - b) = 4d + 1.
In logarithmic form, the base is the subscript and the value inside the logarithm is the argument. So, the equation becomes logc(7a - b) = 4d + 1.
log4(x+1)+log4(4x-3) = 0
log(x+1)/(log4) + log(4x+3)/log4 = 0
times log4
log(x+1) + log(4x+3) = 0
log((x+1)(4x+3)) = 0
10^0 = 4x^2 + 7x + 3
4x^2 + 7x + 2 = 0 , where x > -1
x = (√17 - 7)/8
for the 2nd, there is "no following"