Mix Examples-Logarithms, Indices and Surds, Partial Fractions Questions in English

(C) Given $a^2 + 4b^2 = 12ab$.
Adding $4ab$ to both sides,we get $a^2 + 4b^2 + 4ab = 12ab + 4ab$.
$(a + 2b)^2 = 16ab$.
Taking the logarithm on both sides,we get $\log((a + 2b)^2) = \log(16ab)$.
$2\log(a + 2b) = \log 16 + \log a + \log b$.
Since $16 = 2^4$,$\log 16 = 4\log 2$.
$2\log(a + 2b) = 4\log 2 + \log a + \log b$.
Therefore,$\log(a + 2b) = \frac{1}{2}[\log a + \log b + 4\log 2]$.

(D) Let $\frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b} = k$.
Then,$\log x = k(b - c)$,$\log y = k(c - a)$,and $\log z = k(a - b)$.
For option $A$: $xyz = e^{\log x + \log y + \log z} = e^{k(b - c + c - a + a - b)} = e^0 = 1$.
For option $B$: $x^a y^b z^c = e^{a \log x + b \log y + c \log z} = e^{k(a(b - c) + b(c - a) + c(a - b))} = e^{k(ab - ac + bc - ba + ca - cb)} = e^0 = 1$.
For option $C$: $x^{b + c} y^{c + a} z^{a + b} = e^{(b + c) \log x + (c + a) \log y + (a + b) \log z} = e^{k((b + c)(b - c) + (c + a)(c - a) + (a + b)(a - b))} = e^{k(b^2 - c^2 + c^2 - a^2 + a^2 - b^2)} = e^0 = 1$.
Since all options are true,the correct answer is $D$.

(A) Given: ${a^x} = {(x + y + z)^y}$,${a^y} = {(x + y + z)^z}$,${a^z} = {(x + y + z)^x}$.
Multiplying the three equations,we get:
${a^x} \cdot {a^y} \cdot {a^z} = {(x + y + z)^y} \cdot {(x + y + z)^z} \cdot {(x + y + z)^x}$
${a^{x + y + z}} = {(x + y + z)^{x + y + z}}$
This implies $x + y + z = a$.
Substituting $x + y + z = a$ into the original equations:
${a^x} = {a^y} \Rightarrow x = y$
${a^y} = {a^z} \Rightarrow y = z$
Since $x = y = z$ and $x + y + z = a$,we have $3x = a$,which means $x = y = z = a/3$.

(B) Given equation: $9^x - 2^{x + 1/2} = 2^{x + 3/2} - 3^{2x - 1}$
Rearranging the terms: $3^{2x} + 3^{2x-1} = 2^{x + 3/2} + 2^{x + 1/2}$
Factor out common terms: $3^{2x-1}(3 + 1) = 2^{x + 1/2}(2 + 1)$
$4 \cdot 3^{2x-1} = 3 \cdot 2^{x + 1/2}$
$3^{2x-2} = 2^{x + 1/2 - 2} = 2^{x - 3/2}$
Taking $\log_{(9/2)}$ on both sides: $(x-1) \log_{(9/2)} 9 = (x - 3/2) \log_{(9/2)} 2$
Alternatively,rewrite as: $(9/2)^{x-1} = 2^{x-1} \cdot 3^{2x-2} / 2^{x-1} = (3^2/2)^{x-1} = (9/2)^{x-1} = 2^{-1/2}$
$x - 1 = \log_{(9/2)} (2^{-1/2}) = -1/2 \log_{(9/2)} 2$
$x = 1 - \log_{(9/2)} \sqrt{2} = \log_{(9/2)} (9/2) - \log_{(9/2)} \sqrt{2} = \log_{(9/2)} (9 / (2\sqrt{2})) = \log_{(9/2)} (9/\sqrt{8})$.

(A) Let $x = 0.2343434...$ (Equation $1$)
Multiply by $10$: $10x = 2.343434...$ (Equation $2$)
Multiply Equation $1$ by $1000$: $1000x = 234.343434...$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$1000x - 10x = 234.343434... - 2.343434...$
$990x = 232$
$x = \frac{232}{990}$

(A) Let $x = 0.4232323...$ (Equation $1$)
Multiply by $10$ on both sides: $10x = 4.232323...$ (Equation $2$)
Multiply Equation $1$ by $1000$: $1000x = 423.232323...$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$1000x - 10x = 423.232323... - 4.232323...$
$990x = 419$
$x = \frac{419}{990}$.

(D) The series $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ is an infinite geometric progression with first term $a_1 = \frac{1}{4}$ and common ratio $r = \frac{1}{2}$.
Using the sum formula $S_{\infty} = \frac{a_1}{1 - r}$,we get $x = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.
Now,substitute $a = 0.2 = \frac{1}{5}$,$b = \sqrt{5} = 5^{1/2}$,and $x = \frac{1}{2}$ into the expression $a^{\log_b x}$:
$a^{\log_b x} = (\frac{1}{5})^{\log_{\sqrt{5}} (1/2)} = (5^{-1})^{\log_{5^{1/2}} (2^{-1})}$.
Using the property $\log_{b^n} x = \frac{1}{n} \log_b x$,we have $\log_{5^{1/2}} (2^{-1}) = \frac{1}{1/2} \log_5 (2^{-1}) = 2 \log_5 (2^{-1}) = \log_5 (2^{-2}) = \log_5 (1/4)$.
Thus,the expression becomes $(5^{-1})^{\log_5 (1/4)} = 5^{-\log_5 (1/4)} = 5^{\log_5 (1/4)^{-1}} = 5^{\log_5 4} = 4$.

(C) Let $x = 0.5737373......$
This can be written as $x = 0.5 + 0.0737373......$
$x = \frac{5}{10} + \frac{73}{999} \times \frac{1}{10}$
$x = \frac{1}{2} + \frac{73}{9990}$
$x = \frac{4995 + 73}{9990} = \frac{5068}{9990}$
Alternatively,using the repeating decimal rule:
$x = \frac{573 - 5}{990} = \frac{568}{990}$
Thus,the correct option is $C$.

(D) Given the equation: $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$
For the expression to be defined,we must have $x - 2 > 0$ and $x - 2 \neq 1$,which implies $x > 2$ and $x \neq 3$.
Using the property $\log a + \log b = \log (ab)$,the equation becomes:
$\log (5(x^2 + 1)) = 2 \log (x - 2)$
Using the property $n \log a = \log (a^n)$,we get:
$\log (5x^2 + 5) = \log ((x - 2)^2)$
Equating the arguments:
$5x^2 + 5 = x^2 - 4x + 4$
$4x^2 + 4x + 1 = 0$
$(2x + 1)^2 = 0$
$x = -\frac{1}{2}$
Since the domain condition is $x > 2$,the value $x = -\frac{1}{2}$ is not a valid solution.
Therefore,the number of solutions is $0$.

(B) Given equation: $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$
Rearranging the terms:
$2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$
Factor out the common terms:
$2^{2x - 1}(2 + 1) = 3^{x - 1/2}(3 + 1)$
$2^{2x - 1}(3) = 3^{x - 1/2}(4)$
Divide both sides by $3$ and $3^{x - 1/2}$:
$\frac{2^{2x - 1}}{4} = \frac{3^{x - 1/2}}{3}$
$2^{2x - 1 - 2} = 3^{x - 1/2 - 1}$
$2^{2x - 3} = 3^{x - 3/2}$
Rewrite the exponent on the left side:
$2^{2(x - 3/2)} = 3^{x - 3/2}$
$(2^2)^{x - 3/2} = 3^{x - 3/2}$
$4^{x - 3/2} = 3^{x - 3/2}$
Divide by $3^{x - 3/2}$:
$\left(\frac{4}{3}\right)^{x - 3/2} = 1$
Since $a^0 = 1$,we have:
$x - 3/2 = 0$
$x = 3/2$

(C) Since $a_n > 1$ for all $n \in N$,all logarithmic terms are positive.
Using the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for $n$ positive terms:
$\frac{\log_{a_2} a_1 + \log_{a_3} a_2 + \dots + \log_{a_1} a_n}{n} \ge \sqrt[n]{(\log_{a_2} a_1) \cdot (\log_{a_3} a_2) \cdot \dots \cdot (\log_{a_1} a_n)}$
Using the change of base formula $\log_b a = \frac{\ln a}{\ln b}$,the product becomes:
$\frac{\ln a_1}{\ln a_2} \cdot \frac{\ln a_2}{\ln a_3} \cdot \dots \cdot \frac{\ln a_n}{\ln a_1} = 1$
Therefore,$\frac{\sum \log_{a_{i+1}} a_i}{n} \ge \sqrt[n]{1} = 1$
$\sum \log_{a_{i+1}} a_i \ge n$
Thus,the minimum value is $n$.

(B) Given the equation: $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$.
Rearranging the terms: $2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$.
Factor out the common terms: $2^{2x - 1}(2 + 1) = 3^{x - 1/2}(3 + 1)$.
Simplify: $2^{2x - 1} \cdot 3 = 3^{x - 1/2} \cdot 4$.
$2^{2x - 1} \cdot 3 = 3^{x - 1/2} \cdot 2^2$.
$2^{2x - 3} = 3^{x - 1/2 - 1} = 3^{x - 3/2}$.
Taking $\log$ on both sides: $(2x - 3) \log 2 = (x - 3/2) \log 3$.
$(2x - 3) \log 2 = \frac{2x - 3}{2} \log 3$.
$(2x - 3) \log 2 = (2x - 3) \log \sqrt{3}$.
This implies $2x - 3 = 0$ or $\log 2 = \log \sqrt{3}$ (which is impossible).
Therefore,$2x - 3 = 0$,which gives $x = 3/2$.

(D) We simplify each term individually using the properties of logarithms: $\log_a b = 1/\log_b a$ and $a^{\log_a x} = x$.
Term $1$: ${81^{(1/{\log_5}3)}} = {81^{\log_3 5}} = {(3^4)^{\log_3 5}} = {3^{4 \log_3 5}} = {3^{\log_3 5^4}} = {5^4} = 625$.
Term $2$: ${27^{\log_9 36}} = {(3^3)^{\log_{3^2} 36}} = {3^{3 \cdot \frac{1}{2} \log_3 36}} = {3^{\frac{3}{2} \log_3 36}} = {3^{\log_3 36^{3/2}}} = {36^{3/2}} = {(6^2)^{3/2}} = 6^3 = 216$.
Term $3$: ${3^{4/{\log_7}9}} = {3^{4 \log_9 7}} = {3^{4 \log_{3^2} 7}} = {3^{4 \cdot \frac{1}{2} \log_3 7}} = {3^{2 \log_3 7}} = {3^{\log_3 7^2}} = 7^2 = 49$.
Summing the terms: $625 + 216 + 49 = 890$.

(A) Given ${2^x} = {4^y} = {8^z}$.
Writing in terms of base $2$,we get ${2^x} = {2^{2y}} = {2^{3z}}$.
Equating the exponents,$x = 2y = 3z = k$ (let).
Then $x = k$,$y = k/2$,and $z = k/3$.
Given $xyz = 288$,so $k \times (k/2) \times (k/3) = 288$.
$k^3 / 6 = 288 \implies k^3 = 1728 \implies k = 12$.
Thus,$x = 12$,$y = 6$,and $z = 4$.
Now,calculate $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = \frac{1}{2(12)} + \frac{1}{4(6)} + \frac{1}{8(4)}$.
$= \frac{1}{24} + \frac{1}{24} + \frac{1}{32} = \frac{2}{24} + \frac{1}{32} = \frac{1}{12} + \frac{1}{32}$.
$= \frac{8 + 3}{96} = \frac{11}{96}$.

(B) Given that ${a^x} = {b^y} = {(ab)^{xy}}$.
Let ${a^x} = {b^y} = {(ab)^{xy}} = k$.
Then $a^x = k \implies a = k^{1/x}$ and $b^y = k \implies b = k^{1/y}$.
Also,$(ab)^{xy} = k \implies ab = k^{1/xy}$.
Substituting the values of $a$ and $b$ in the last equation:
$k^{1/x} \cdot k^{1/y} = k^{1/xy}$
$k^{(1/x + 1/y)} = k^{1/xy}$
Equating the exponents:
$\frac{1}{x} + \frac{1}{y} = \frac{1}{xy}$
$\frac{x+y}{xy} = \frac{1}{xy}$
Since $x, y \neq 0$,we get $x + y = 1$.

(D) Let $x = 4 + \sqrt{15}$ and $y = 4 - \sqrt{15}$. Then $x+y = 8$ and $xy = 16 - 15 = 1$.
We can write $x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{2x} + \sqrt{2y})^3$.
Since $2x = 8 + 2\sqrt{15} = (\sqrt{5} + \sqrt{3})^2$ and $2y = (\sqrt{5} - \sqrt{3})^2$,we have $\sqrt{2x} = \sqrt{5} + \sqrt{3}$ and $\sqrt{2y} = \sqrt{5} - \sqrt{3}$.
Thus,$x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{5} + \sqrt{3} + \sqrt{5} - \sqrt{3})^3 = \frac{1}{\sqrt{2}} (2\sqrt{5})^3 = \frac{40\sqrt{5}}{\sqrt{2}} = 20\sqrt{10}$.
Similarly,for the denominator,let $a = 6 + \sqrt{35}$ and $b = 6 - \sqrt{35}$. Then $2a = 12 + 2\sqrt{35} = (\sqrt{7} + \sqrt{5})^2$ and $2b = (\sqrt{7} - \sqrt{5})^2$.
$a^{3/2} - b^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{2a}^3 - \sqrt{2b}^3) = \frac{1}{\sqrt{2}} ((\sqrt{7} + \sqrt{5})^3 - (\sqrt{7} - \sqrt{5})^3)$.
Using $(u+v)^3 - (u-v)^3 = 2v^3 + 6u^2v$,we get $\frac{1}{\sqrt{2}} (2(5\sqrt{5}) + 6(7)\sqrt{5}) = \frac{1}{\sqrt{2}} (10\sqrt{5} + 42\sqrt{5}) = \frac{52\sqrt{5}}{\sqrt{2}} = 26\sqrt{10}$.
The ratio is $\frac{20\sqrt{10}}{26\sqrt{10}} = \frac{20}{26} = \frac{10}{13}$.

(B) Given $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.
First,calculate $x + y$ and $xy$:
$x + y = \frac{(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5} - \sqrt{2})^2}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{(5 + 2 + 2\sqrt{10}) + (5 + 2 - 2\sqrt{10})}{5 - 2} = \frac{14}{3}$.
$xy = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} = 1$.
Now,$3x^2 - 3y^2 + 4xy = 3(x - y)(x + y) + 4xy$.
We know $(x - y)^2 = (x + y)^2 - 4xy = (\frac{14}{3})^2 - 4(1) = \frac{196}{9} - 4 = \frac{196 - 36}{9} = \frac{160}{9}$.
So,$x - y = \frac{\sqrt{160}}{3} = \frac{4\sqrt{10}}{3}$.
Substituting these values into the expression:
$3(\frac{4\sqrt{10}}{3})(\frac{14}{3}) + 4(1) = \frac{56\sqrt{10}}{3} + 4 = \frac{56\sqrt{10} + 12}{3} = \frac{1}{3}[56\sqrt{10} + 12]$.

(C) Let $y = 0.c\overline{ab}$.
This can be written as $y = \frac{cab - c}{990}$.
Expanding the digits,we have $cab = 100c + 10a + b$.
Substituting this into the expression,we get $y = \frac{(100c + 10a + b) - c}{990}$.
Simplifying the numerator,$y = \frac{99c + 10a + b}{990}$.

(B) Given $x \in \left(0, \frac{\pi}{2}\right)$.
From the first equation: $\log_{10} \sin x + \log_{10} \cos x = -1$
$\Rightarrow \log_{10}(\sin x \cos x) = -1$
$\Rightarrow \sin x \cos x = 10^{-1} = \frac{1}{10} \quad ....(1)$
From the second equation: $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) = \frac{1}{2} \log_{10} \left(\frac{n}{10}\right) = \log_{10} \sqrt{\frac{n}{10}}$
$\Rightarrow \sin x + \cos x = \sqrt{\frac{n}{10}}$
Squaring both sides:
$(\sin x + \cos x)^2 = \frac{n}{10}$
$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{n}{10}$
$1 + 2 \left(\frac{1}{10}\right) = \frac{n}{10}$
$1 + \frac{1}{5} = \frac{n}{10}$
$\frac{6}{5} = \frac{n}{10}$
$n = \frac{6 \times 10}{5} = 12$.

(B) Given equations are:
$1) \log _{1 / 3}(x+y)+\log _3(x-y)=2$
$2) 2^{y^2}=512^{x+1}$
From equation $(1)$:
$-\log _3(x+y)+\log _3(x-y)=2$
$\log _3\left(\frac{x-y}{x+y}\right)=2$
$\frac{x-y}{x+y}=3^2=9$
$x-y=9x+9y$
$-8x=10y \Rightarrow y = -\frac{4}{5}x$
From equation $(2)$:
$2^{y^2}=(2^9)^{x+1} = 2^{9(x+1)}$
$y^2=9(x+1)$
Substitute $y = -\frac{4}{5}x$ into $y^2=9(x+1)$:
$\left(-\frac{4}{5}x\right)^2=9(x+1)$
$\frac{16}{25}x^2=9x+9$
$16x^2=225x+225$
$16x^2-225x-225=0$
$(16x+15)(x-15)=0$
So,$x=15$ or $x=-\frac{15}{16}$.
For $x=15$,$y=-\frac{4}{5}(15)=-12$. Check domain: $x+y=15-12=3 > 0$ and $x-y=15-(-12)=27 > 0$. This is a valid solution.
For $x=-\frac{15}{16}$,$y=-\frac{4}{5}(-\frac{15}{16})=\frac{3}{4}$. Check domain: $x+y=-\frac{15}{16}+\frac{12}{16}=-\frac{3}{16} < 0$. This is not valid as the argument of the logarithm must be positive.
Thus,there is only $1$ solution pair $(15, -12)$.

(C) Given the equation $2^a + 2^{2b} + 2^{3c} = 328$.
Since $328 = 256 + 64 + 8 = 2^8 + 2^6 + 2^3$,we test powers of $2$.
We observe that $328 = 2^3 + 2^6 + 2^8$.
Comparing $2^a + 2^{2b} + 2^{3c} = 2^3 + 2^6 + 2^8$,we can set $a=3$,$2b=6 \Rightarrow b=3$,and $3c=8$ (which is not an integer).
Alternatively,let us re-evaluate: $2^a + 2^{2b} + 2^{3c} = 328$.
If $c=1$,$2^a + 2^{2b} = 328 - 8 = 320$. $2^a(1 + 2^{2b-a}) = 320 = 2^6 \times 5$. So $a=6$,$1 + 2^{2b-6} = 5$ $\Rightarrow 2^{2b-6} = 4 = 2^2$ $\Rightarrow 2b-6=2$ $\Rightarrow b=4$.
Thus,$(a, b, c) = (6, 4, 1)$.
Check: $2^6 + 4^4 + 8^1 = 64 + 256 + 8 = 328$. This works.
Now calculate $\frac{a + 2b + 3c}{abc} = \frac{6 + 2(4) + 3(1)}{6 \times 4 \times 1} = \frac{6 + 8 + 3}{24} = \frac{17}{24}$.

(B) Let $x = 0.75$.
The given expression is $\frac{x^3}{1-x} + (x + x^2 + 1)$.
We know that $(1-x)(1+x+x^2) = 1-x^3$.
So,the expression becomes $\frac{x^3 + (1-x)(1+x+x^2)}{1-x} = \frac{x^3 + 1 - x^3}{1-x} = \frac{1}{1-x}$.
Substituting $x = 0.75$:
$\frac{1}{1-0.75} = \frac{1}{0.25} = 4$.
The square root of the result is $\sqrt{4} = 2$.

(D) Given the equation: $3^{(x/y)+1} - 3^{(x/y)-1} = 24$
Let $u = x/y$. Then the equation becomes $3^{u+1} - 3^{u-1} = 24$.
Factor out $3^{u-1}$:
$3^{u-1} \cdot (3^2 - 1) = 24$
$3^{u-1} \cdot (9 - 1) = 24$
$3^{u-1} \cdot 8 = 24$
$3^{u-1} = 3$
Since $3^{u-1} = 3^1$,we have $u - 1 = 1$,which implies $u = 2$.
Thus,$x/y = 2$.
We need to find the value of $\frac{x+y}{x-y}$. Dividing the numerator and denominator by $y$:
$\frac{x/y + 1}{x/y - 1} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3$.

(A) Given $p = 99$ and $q = 101$.
$p_1 = \log_{10} \left(\frac{99+101}{2}\right) = \log_{10} 100 = 2$.
$q_1 = \frac{1}{2}(\log_{10} 99 + \log_{10} 101) = \log_{10} \sqrt{99 \times 101} = \log_{10} \sqrt{9999}$.
Since $9999 < 10000$,$\sqrt{9999} < 100$,so $q_1 < \log_{10} 100 = 2$.
Thus,$p_1 > q_1$.
By the Arithmetic Mean-Geometric Mean inequality,for any two positive numbers $a$ and $b$,$\frac{a+b}{2} > \sqrt{ab}$ if $a \neq b$.
Since $p_1 > q_1$,we have $\frac{p_1+q_1}{2} > \sqrt{p_1 q_1}$.
Taking $\log_{10}$ on both sides,$\log_{10} \left(\frac{p_1+q_1}{2}\right) > \log_{10} \sqrt{p_1 q_1} = \frac{1}{2}(\log_{10} p_1 + \log_{10} q_1)$.
This implies $p_2 > q_2$.
Also,since $p_1 > q_1$,it follows that $\log_{10} p_1 > \log_{10} q_1$.
Since $p_1 > \frac{p_1+q_1}{2} > q_1$,applying the logarithm function (which is strictly increasing) gives $\log_{10} p_1 > p_2 > \log_{10} q_1$.
Combining these,we get $\log_{10} p_1 > p_2 > q_2 > \log_{10} q_1$.

(C) Let $x = 1.\overline{41} = 1.414141...$ $(i)$
Multiply both sides by $100$ to shift the decimal point:
$100x = 141.414141...$ (ii)
Subtract equation $(i)$ from equation (ii):
$100x - x = 141.414141... - 1.414141...$
$99x = 140$
$x = \frac{140}{99}$
Therefore,the rational form is $\frac{140}{99}$.

(D) Given that $a, b, c$ are sides of a right-angled triangle with $c$ as the hypotenuse,we have $a^2 + b^2 = c^2$,which implies $a^2 = c^2 - b^2 = (c+b)(c-b)$.
Taking the logarithm on both sides,we get $\log(a^2) = \log((c+b)(c-b)) = \log(c+b) + \log(c-b)$.
Now,the given expression is $E = \frac{\log_{c+b} a + \log_{c-b} a}{2 \log_{c+b} a \cdot \log_{c-b} a}$.
Using the change of base formula $\log_x y = \frac{\log y}{\log x}$,we have:
$E = \frac{\frac{\log a}{\log(c+b)} + \frac{\log a}{\log(c-b)}}{2 \cdot \frac{\log a}{\log(c+b)} \cdot \frac{\log a}{\log(c-b)}}$.
Simplifying the numerator: $\frac{\log a (\log(c-b) + \log(c+b))}{\log(c+b) \log(c-b)}$.
Simplifying the denominator: $\frac{2 (\log a)^2}{\log(c+b) \log(c-b)}$.
Thus,$E = \frac{\log a (\log((c+b)(c-b)))}{2 (\log a)^2} = \frac{\log a \cdot \log(a^2)}{2 (\log a)^2} = \frac{\log a \cdot 2 \log a}{2 (\log a)^2} = \frac{2 (\log a)^2}{2 (\log a)^2} = 1$.

(B) The given equation is $\log_{(x+1)}((x+1)(2x+3)) = 4 - \log_{(2x+3)}(x+1)^2$.
Using the property $\log_a(bc) = \log_a b + \log_a c$,we get $1 + \log_{(x+1)}(2x+3) = 4 - 2 \log_{(2x+3)}(x+1)$.
Let $y = \log_{(x+1)}(2x+3)$. Then $\log_{(2x+3)}(x+1) = 1/y$.
The equation becomes $1 + y = 4 - 2/y$.
Multiplying by $y$ (where $y \neq 0$),we get $y + y^2 = 4y - 2$,which simplifies to $y^2 - 3y + 2 = 0$.
Factoring gives $(y-1)(y-2) = 0$,so $y=1$ or $y=2$.
Case $1$: $\log_{(x+1)}(2x+3) = 1 \implies x+1 = 2x+3 \implies x = -2$. Checking the base constraints $(x+1 > 0, x+1 \neq 1)$,$x = -2$ is invalid.
Case $2$: $\log_{(x+1)}(2x+3) = 2 \implies (x+1)^2 = 2x+3 \implies x^2 + 2x + 1 = 2x+3 \implies x^2 = 2 \implies x = \pm \sqrt{2}$.
Checking constraints: For $x = -\sqrt{2}$,the base $x+1 = 1-\sqrt{2} < 0$,which is invalid. For $x = \sqrt{2}$,the bases $x+1 = 1+\sqrt{2} > 0$ and $2x+3 = 2\sqrt{2}+3 > 0$ are valid.
Thus,the only real solution is $x = \sqrt{2}$.
The sum of squares of all real solutions is $(\sqrt{2})^2 = 2$.