Algebra Questions Answers

We have log $a$ = $x$ log $b$
log $b$ = $y$ log $c$ and
loc $c$ = $z$ log $a$
 $therefore$ log $a$ log $b$ log $c$
= $xyz$ log $a$ log $b$
log $c$  $Rightarrow xyz = 1$

Given, $5x + 10y = 40$ i.e., $x + 2y = 8$ and $x + y = 7$.
After solving both the equations, we have $x = 6$ and $y = 1$

$= 1 + frac{x}{27} = sqrt{frac{784}{729}} = frac{28}{27} = 1 + frac{1}{27}$

Hence the  value of $x$ = 1

$text{R } = fleft(-3right) = 5left(-3right)^3 + 5left(-3right)^2 - 6(-3) + 9$

$= - 63$

$begin{aligned} & x^4 + y^4 - x^3y^2 - x^2y^3 + 16xy \ & [(x + y)^2 - 2xy]^2 - 4x^2y^2 + 16xy \ & (x + y)^4 - 4(x + y)^2 xy + 4x^2y^2 - 4x^2y^2 + 16xy \ & (x + y = 2 text{ given}) \ & (x + y)^4 = 2^4 = 2 times 2 times 2 times 2 = 16 end{aligned}$

$=alpha + beta = - frac{2b}{a}, alphabeta = frac{c}{a}$

$=sqrt{frac{alpha}{beta}} + sqrt{frac{beta}{alpha}} = frac{alpha + beta}{sqrt {alphabeta}} = frac{-frac{2b}{a}}{sqrt{frac{c}{a}}} = frac{-2b}{sqrt{ac}}$

$2^x = 2^{2y} = 2^{3z}$

$x = 2y =3z$

$(3z) biggl(frac{3}{2}z biggr)z = 288$

$text{ or } z^3 = frac{288 times 2}{9} = frac{576}{9} = 64$

$z = 4, x = 12, y = 6$

$frac{1}{2x} + frac{1}{4y} + frac{1}{8z} = frac{1}{24} + frac{1}{24} + frac{1}{24} = frac{3}{24} = 8$

$biggl(frac{a}{b}biggr)^{x-1} = biggl(frac{b}{a}biggr)^{x-3}$

$text{or }biggl(frac{a}{b}biggr)^{x-1} = biggl(frac{a}{b}biggr)^{-x-3}$

$therefore quad x-1 = -x + 3$

$2x = 4$

$x = 2$

$begin{aligned} 2^{x-7} &= 256 &= 2^8 therefore x-7 &= 8 text{or } x &= 15 end{aligned}$

By question, $x propto frac{1}{y^2} Rightarrow x = k quad frac{1}{y^2}$

[where $k$ is a constant]

$Rightarrow xy^2 = k$

Here $y = 2$ for $x = 1$

then $1 times 4 = k$

$Rightarrow k = 4$

Now, if  $y = 6 quad text{ then } quad x times 36 = 4$
$Rightarrow x = frac{1}{9}$