1. solve the following logarithmic equation log_8(x+8)+log_8(x+7)=2
what is the exact solution
2.se the rational theorem to find all the real zeros of the polynomial function. use the zeros to factor f over the real numbers.
f(x)=4x^4+5x^3+9x^2+10x+2
a.find the real zeros of f. x=?
b.then use the real zeros to factor f. f(x)=?
please show work.
using rules of logs, your equation becomes
log8 ((x+8)/(x+7)) = 2
(x+8)(x+7) = 8^2 = 64
x^2 + 15x + 56 - 64 = 0
x^2 + 15x - 8 = 0
x = (-15 ± √257)/2
but x > -7 or else the log part would be undefined
so x = (-15 + √257)/2
2.
if rational roots exist, they must be ±1, or ±2 or ±1/2, ±1/4
we can rule out any positive x's since all the terms have positive and will never add up to zero
f(-1) = 4 - 5 + 9 - 10 + 2 = 0 , yeahh
by synthetic division
4x^4+5x^3+9x^2+10x+2
= (x+1)(4x^3 + x^2 + 8x + 2) , again all positives in the 2nd factor so try x = -1, -2, -1/2, -1/4
x=-1
f(-1) = -4+1 -8 + 2 ≠ 0
f(-2) = -32 + ... ≠0
...
f(-1/4) = ... 0
so (4x+1) is a factor
Using long algebraic division I ended up with
4x^4+5x^3+9x^2+10x+2 = (x+1)(4x+1)(x^2 + 2)
so there are 2 real roots or zeros:
x = -1 and x = -1/4
1. To solve the logarithmic equation log_8(x+8) + log_8(x+7) = 2, we can use the logarithmic properties:
The sum of logarithms with the same base is equal to the logarithm of the product. Therefore, we can rewrite the equation as a single logarithmic expression:
log_8((x+8)(x+7)) = 2
Next, we can rewrite the logarithmic equation in its exponential form:
8^2 = (x+8)(x+7)
64 = (x+8)(x+7)
Now we have a quadratic equation. We can expand it and rewrite it in standard form:
64 = x^2 + 15x + 56
Rearranging the terms, we get:
x^2 + 15x + 56 - 64 = 0
x^2 + 15x - 8 = 0
Now we can solve this equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a
For our equation, a = 1, b = 15, and c = -8. Plugging in these values, we get:
x = [-15 ± √(15^2 - 4(1)(-8))] / 2(1)
x = [-15 ± √(225 + 32)] / 2
x = [-15 ± √(257)] / 2
Therefore, the exact solutions to the logarithmic equation are:
x = (-15 + √(257)) / 2
x = (-15 - √(257)) / 2
2. To find all the real zeros of the polynomial function f(x) = 4x^4 + 5x^3 + 9x^2 + 10x + 2, we can use the rational root theorem.
a. According to the rational root theorem, any rational zero of a polynomial can be expressed as p/q, where p is a factor of the constant term (in this case, 2), and q is a factor of the leading coefficient (which is 4).
The factors of 2 are ±1 and ±2, and the factors of 4 are ±1 and ±4. Therefore, the possible rational roots are ±1, ±2, ±1/2, and ±1/4.
We can test these possible roots by plugging them into the polynomial and checking if the result is zero:
f(1) = 4(1)^4 + 5(1)^3 + 9(1)^2 + 10(1) + 2 = 30 ≠ 0
f(-1) = 4(-1)^4 + 5(-1)^3 + 9(-1)^2 + 10(-1) + 2 = 2 ≠ 0
f(2) = 4(2)^4 + 5(2)^3 + 9(2)^2 + 10(2) + 2 = 114 ≠ 0
f(-2) = 4(-2)^4 + 5(-2)^3 + 9(-2)^2 + 10(-2) + 2 = 230 ≠ 0
f(1/2) = 4(1/2)^4 + 5(1/2)^3 + 9(1/2)^2 + 10(1/2) + 2 = 9.5 ≠ 0
f(-1/2) = 4(-1/2)^4 + 5(-1/2)^3 + 9(-1/2)^2 + 10(-1/2) + 2 = 7.75 ≠ 0
f(1/4) = 4(1/4)^4 + 5(1/4)^3 + 9(1/4)^2 + 10(1/4) + 2 = 6.0625 ≠ 0
f(-1/4) = 4(-1/4)^4 + 5(-1/4)^3 + 9(-1/4)^2 + 10(-1/4) + 2 = 2.5625 ≠ 0
None of these values are zero, which means there are no rational zeros (real zeros that are fractions). However, there may still be irrational or complex zeros.
b. To factor f(x) over the real numbers, we can use the information from part (a) to conclude that the polynomial has no linear factors. Therefore, the polynomial cannot be factored further over the real numbers, as it is already in its factored form:
f(x) = 4x^4 + 5x^3 + 9x^2 + 10x + 2