Math Snap

STEP 1

Assumptions1. A binary operation $*$ is defined on $\mathrm{R}$, the set of real numbers by $a * b=a+b-3$.
. The sequence is $4,-,1,-\frac{1}{}, \ldots$.
3. The equation to solve is $256^{(x+1)}=8^{\left(1-x^{}\right)}$.
4. The roots of the equation $x^{}+3 x-4=0$ are $\alpha$ and $\beta$.
5. The expression to rationalize is $\frac{1}{\sqrt{3}-\sqrt{}}$.
6. The equation to solve is $\log{5}(6 x+7)-\log{5}6=$.

STEP 2

For the first problem, the identity element, $e$, is defined such that $a <em> e = a$ for all $a$ in the set. So we set up the equation $a </em> e = a$ and solve for $e$.
$$a * e = a$$$$a + e - = a$$

STEP 3

olving the equation for $e$ gives$$e -3 =0$$$$e =3$$So, the identity element for the operation $*$ is3. Answer A.3

STEP 4

For the second problem, the sequence is a geometric sequence with first term $a =4$ and common ratio $r = -1/2$. The sum of an infinite geometric sequence is given by $= a / (1 - r)$.
$$= \frac{4}{1 - (-1/2)}$$

STEP 5

olving the equation for $$gives$$ = \frac{4}{1.5} = \frac{8}{3}$$So, the sum of the sequence is $\frac{8}{3}$. Answer C. $\frac{8}{3}$

STEP 6

For the third problem, we can rewrite the equation $256^{(x+1)}=8^{\left(1-x^{2}\right)}$ as $2^{8(x+1)} =2^{3(1-x^{2})}$.

STEP 7

Since the bases are equal, we can set the exponents equal to each other and solve for $x$.
$$(x+1) =3(1-x^{2})$$

STEP 8

olving the equation for $x$ gives$$x = \frac{8}{3}, \frac{5}{3}$$So, the solutions to the equation are $\frac{8}{3}$ and $\frac{5}{3}$. Answer D. $\frac{8}{3}, \frac{5}{3}$

STEP 9

For the fourth problem, we know that the sum of the roots of the equation $x^{2}+3 x-4=$ is $-\frac{b}{a} = -3$ and the product of the roots is $\frac{c}{a} = -4$. We can use these to find the value of $\alpha^{2}+\beta^{2}-3 \alpha \beta$.
$$\alpha^{2}+\beta^{2}-3 \alpha \beta = (\alpha+\beta)^{2} -3\alpha\beta -4\alpha\beta$$

STEP 10

Substituting the values of $\alpha+\beta$ and $\alpha\beta$ gives$$\alpha^{2}+\beta^{2}-3 \alpha \beta = (-3)^{2} -3(-4) -4(-4) =9 +12 +16 =37$$So, the value of $\alpha^{2}+\beta^{2}-3 \alpha \beta$ is37. This is not an option in the given choices, so there seems to be a mistake in the problem or the choices.

STEP 11

For the fifth problem, we can rationalize the denominator of the expression $\frac{}{\sqrt{3}-\sqrt{}}$ by multiplying the numerator and denominator by the conjugate of the denominator.
$$\frac{}{\sqrt{3}-\sqrt{}} \times \frac{\sqrt{3}+\sqrt{}}{\sqrt{3}+\sqrt{}} = \frac{\sqrt{3}+\sqrt{}}{3-} = \sqrt{3}+\sqrt{}$$So, the rationalized form of the expression is $\sqrt{3}+\sqrt{}$. Answer D. $\sqrt{3}+\sqrt{}$

STEP 12

For the sixth problem, we can use the properties of logarithms to solve the equation $\log{5}(6 x+7)-\log{5}6=2$.
$$\log{5}\left(\frac{6x+7}{6}\right) =2$$$$\frac{6x+7}{6} =5^{2}$$

olving the equation for $x$ gives$$x = \frac{25 \times6 -7}{6} =19$$So, the solution to the equation is $x =19$. This is not an option in the given choices, so there seems to be a mistake in the problem or the choices.