Logarithms Questions in English

(C) Given equations are:
$1$) $\log _3 x + \log _3 y = 2 + \log _3 2$
$2$) $\log _3(x+y) = 2$
From equation $(1)$,using the property $\log_a b + \log_a c = \log_a(bc)$:
$\log _3(xy) = \log _3(3^2) + \log _3 2$
$\log _3(xy) = \log _3 9 + \log _3 2$
$\log _3(xy) = \log _3(9 \times 2) = \log _3 18$
So,$xy = 18$.
From equation $(2)$,using the definition of logarithm:
$x + y = 3^2 = 9$.
We have a system of equations:
$x + y = 9$
$xy = 18$
These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$:
$t^2 - 9t + 18 = 0$
$(t-3)(t-6) = 0$
So,$t = 3$ or $t = 6$.
Thus,the solutions are $(x=3, y=6)$ or $(x=6, y=3)$.
Comparing with the given options,the correct choice is $x=3, y=6$.

(A) Given that $\log _{7} 2 = \lambda$.
We need to find the value of $\log _{49} (28)$.
Using the change of base formula $\log _{a^n} b = \frac{1}{n} \log _{a} b$,we have:
$\log _{49} (28) = \log _{7^2} (28) = \frac{1}{2} \log _{7} (28)$.
Since $28 = 4 \times 7 = 2^2 \times 7$,we can write:
$\frac{1}{2} \log _{7} (2^2 \times 7) = \frac{1}{2} [\log _{7} (2^2) + \log _{7} 7]$.
Using the property $\log _{a} (x^n) = n \log _{a} x$ and $\log _{a} a = 1$:
$\frac{1}{2} [2 \log _{7} 2 + 1] = \frac{1}{2} (2 \lambda + 1)$.

(D) Using the change of base formula $\log_{a} b = \frac{\log b}{\log a}$:
Numerator = $\log_{3} 5 \times \log_{25} 27 \times \log_{49} 7$
$= \frac{\log 5}{\log 3} \times \frac{\log 27}{\log 25} \times \frac{\log 7}{\log 49}$
$= \frac{\log 5}{\log 3} \times \frac{3 \log 3}{2 \log 5} \times \frac{\log 7}{2 \log 7}$
$= \frac{1}{1} \times \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$
Denominator = $\log_{81} 3 = \log_{3^4} 3 = \frac{1}{4} \log_{3} 3 = \frac{1}{4}$
Value = $\frac{3/4}{1/4} = 3$

(C) Using the property $\frac{1}{\log _{a} b} = \log _{b} a$,we can rewrite the expression as:
$\log _{12} 3 + \log _{12} 4$
Applying the logarithmic property $\log _{b} x + \log _{b} y = \log _{b} (xy)$:
$\log _{12} (3 \times 4) = \log _{12} 12$
Since $\log _{b} b = 1$,the value is $1$.

(B) Given $x = \log_a (bc)$,$y = \log_b (ca)$,and $z = \log_c (ab)$.
Adding $1$ to each term:
$1+x = 1 + \log_a (bc) = \log_a a + \log_a (bc) = \log_a (abc)$.
Similarly,$1+y = \log_b (abc)$ and $1+z = \log_c (abc)$.
Now,taking the reciprocal:
$\frac{1}{1+x} = \frac{1}{\log_a (abc)} = \log_{abc} a$.
$\frac{1}{1+y} = \frac{1}{\log_b (abc)} = \log_{abc} b$.
$\frac{1}{1+z} = \frac{1}{\log_c (abc)} = \log_{abc} c$.
Adding these expressions:
$\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = \log_{abc} a + \log_{abc} b + \log_{abc} c = \log_{abc} (abc) = 1$.

(B) The domain of the expression is $x > 0$.
Given inequality: $\log _{25} x^2 + (\log _5 x)^2 < 2$.
Using the property $\log _{a^n} b^m = \frac{m}{n} \log _a b$,we have $\log _{25} x^2 = \log _{5^2} x^2 = \frac{2}{2} \log _5 x = \log _5 x$.
Substituting this into the inequality: $\log _5 x + (\log _5 x)^2 < 2$.
Let $y = \log _5 x$. Then $y^2 + y - 2 < 0$.
Factoring the quadratic: $(y + 2)(y - 1) < 0$.
This implies $-2 < y < 1$.
Substituting back $y = \log _5 x$: $-2 < \log _5 x < 1$.
Converting to exponential form: $5^{-2} < x < 5^1$.
Thus,$\frac{1}{25} < x < 5$.

(A) For the logarithmic inequality $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$ to be defined,the arguments must be positive:
$1) \ x^{2}-16 > 0$ $\Rightarrow (x-4)(x+4) > 0$ $\Rightarrow x \in (-\infty, -4) \cup (4, \infty)$
$2) \ 4x-11 > 0 \Rightarrow x > \frac{11}{4} = 2.75$
Combining these,the domain is $x > 4$.
Now,solving the inequality:
$x^{2}-16 \leq 4x-11$
$x^{2}-4x-5 \leq 0$
$(x-5)(x+1) \leq 0$
This holds for $x \in [-1, 5]$.
Taking the intersection of the domain $(x > 4)$ and the solution $(x \in [-1, 5])$,we get $4 < x \leq 5$.

(C) Given that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $AP$.
Therefore,$2 \log _{b} x = \log _{a} x + \log _{c} x$.
Using the change of base formula $\log _{m} n = \frac{1}{\log _{n} m}$,we get:
$\frac{2}{\log _{x} b} = \frac{1}{\log _{x} a} + \frac{1}{\log _{x} c}$.
$\frac{2}{\log _{x} b} = \frac{\log _{x} c + \log _{x} a}{\log _{x} a \cdot \log _{x} c} = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
Rearranging,we have $\frac{2}{\log _{x} b} = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
Since $\frac{1}{\log _{x} a} = \log _{a} x$,we can write $\frac{\log _{x} a}{\log _{x} b} = \log _{b} a$,so $\frac{1}{\log _{x} b} = \log _{b} x$.
Actually,$\frac{2}{\log _{x} b} = 2 \log _{b} x$. Thus,$2 \log _{b} x = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
Using $\log _{x} a \cdot \log _{a} b = \log _{x} b$,we get $\log _{x} b = \log _{x} a \cdot \log _{a} b$.
Substituting this,$2 \log _{b} x = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
This simplifies to $\log _{x} c^2 = \log _{x} (ac)^{\log _{a} b}$.
Thus,$c^2 = (ac)^{\log _{a} b}$.

(D) Given the equation: $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$.
Let $x = \frac{2}{3} \log _{b} a$,$y = \frac{3}{5} \log _{c} b$,and $z = \frac{5}{2} \log _{a} c$.
Note that $x \cdot y \cdot z = (\frac{2}{3} \log _{b} a) \cdot (\frac{3}{5} \log _{c} b) \cdot (\frac{5}{2} \log _{a} c) = (\frac{2}{3} \cdot \frac{3}{5} \cdot \frac{5}{2}) \cdot (\log _{b} a \cdot \log _{c} b \cdot \log _{a} c) = 1 \cdot 1 = 1$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$,for positive real numbers $x, y, z$ with $x+y+z=3$ and $xyz=1$,the equality holds only when $x=y=z=1$.
Thus,$\frac{2}{3} \log _{b} a = 1 \Rightarrow \log _{b} a = \frac{3}{2}$.
Given $b=9$,we have $\log _{9} a = \frac{3}{2}$.
Therefore,$a = 9^{3/2} = (3^2)^{3/2} = 3^3 = 27$.

(A) Given,$\log _{0.3}(x-1) < \log _{0.09}(x-1)$.
For the logarithm to be defined,we must have $x-1 > 0$,which implies $x > 1$.
We can rewrite the inequality as $\log _{0.3}(x-1) < \log _{0.3^{2}}(x-1)$.
Using the property $\log _{a^n} b = \frac{1}{n} \log_a b$,we get $\log _{0.3}(x-1) < \frac{1}{2} \log _{0.3}(x-1)$.
Multiplying by $2$,we get $2 \log _{0.3}(x-1) < \log _{0.3}(x-1)$.
Subtracting $\log _{0.3}(x-1)$ from both sides,we get $\log _{0.3}(x-1) < 0$.
Since the base $0.3 < 1$,the inequality reverses when we remove the logarithm: $x-1 > (0.3)^0$.
$x-1 > 1$.
$x > 2$.
Thus,$x$ lies in the interval $(2, \infty)$.

(C) Given equation: $ \log_{(x+3)}(6x^{2}+28x+30)=5-2\log_{(6x+10)}(x+3)^{2} $
Factorize the argument: $ 6x^{2}+28x+30 = 2(3x^{2}+14x+15) = 2(3x+5)(x+3) $. This does not simplify directly. Let us re-examine: $ 6x^{2}+28x+30 = (6x+10)(x+3) $.
So,$ \log_{(x+3)}[(6x+10)(x+3)] = 5 - 4\log_{(6x+10)}(x+3) $
$ 1 + \log_{(x+3)}(6x+10) = 5 - 4\log_{(6x+10)}(x+3) $
Let $ A = \log_{(x+3)}(6x+10) $. Then $ \log_{(6x+10)}(x+3) = \frac{1}{A} $.
The equation becomes $ 1 + A = 5 - \frac{4}{A} $.
$ A - 4 + \frac{4}{A} = 0 \Rightarrow A^{2} - 4A + 4 = 0 $.
$ (A-2)^{2} = 0 \Rightarrow A = 2 $.
$ \log_{(x+3)}(6x+10) = 2 \Rightarrow 6x+10 = (x+3)^{2} $.
$ 6x+10 = x^{2}+6x+9 \Rightarrow x^{2} = 1 $.
$ x = 1 $ or $ x = -1 $.
Check domain: For $ x=1 $,base $ x+3=4 > 0, \neq 1 $ and argument $ 6(1)^{2}+28(1)+30 = 64 > 0 $. Valid.
For $ x=-1 $,base $ x+3=2 > 0, \neq 1 $ and argument $ 6(-1)^{2}+28(-1)+30 = 8 > 0 $. Valid.
Sum of solutions $= 1 + (-1) = 0$.