Mathematics Test - 3

For given expression to be defined, \(3-x>0 \Rightarrow x<3,1-x>0 \Rightarrow x<1 \Rightarrow(i)\)
\(3+x>0,2 x+1>0 \Rightarrow x>-1 / 2\)
Thus domain is : \(x \in\left(-\frac{1}{2}, 1\right)=D\) (say) We have, \(\log _{4}(3-x)+\log _{25}(3+x)=\log _{4}(1-x)+\log _{0.25}(2 x+1)\)
\(\Rightarrow \log _{4}(3-x)+\log _{1 / 4}(3+x)=\log _{4}(1-x)+\log _{1 / 4}(2 x+1)\)
\(\Rightarrow \log _{4}(3-x)-\log _{4}(3+x)=\log _{4}(1-x)-\log _{4}(2 x+1),\left[\log _{1 / a} b=-\log _{a} b\right]\)
\(\Rightarrow \log _{4} \frac{(3-x)}{(3+x)}=\log _{4} \frac{(1-x)}{(2 x+1)},\left[\because \log a-\log b=\log \frac{a}{b}\right]\)
\(\Rightarrow\left(\frac{3-x}{3+x}\right)=\left(\frac{1-x}{2 x+1}\right)\)
\(\Rightarrow 6 x+3-2 x^{2}-x=3+x-3 x-x^{2}\)
\(\Rightarrow x^{2}-7 x=0 \Rightarrow x=0,7\)
But \(7 \notin D, \therefore x=0\) is the only solution Thus, it has only one real solutions.

\begin{equation}\begin{array}{l}
\text { Given, } \log _{e}\left(\frac{a+b}{2}\right)=\frac{1}{2}\left(\log _{e} a+\log _{e} b\right)=\frac{1}{2} \log _{e}(a b),[\because \log x+\log y=\log (x y)] \\
\Rightarrow \log _{e}\left(\frac{a+b}{2}\right)=\left(\log _{e} \sqrt{a b}\right)\left[\because x \log a=\log a^{x}\right] \\
\Rightarrow \frac{a+b}{2}=\sqrt{a b} \\
\Rightarrow a+b-2 \sqrt{a b}=0 \\
\Rightarrow(\sqrt{a}-\sqrt{b})^{2}=0 \\
\Rightarrow a=b \\
\text { Given } a=4 \Rightarrow b=4
\end{array}\end{equation}

Given, \(3=k \cdot 2^{r}\) and \(15=k .4^{r}\)
We have, \(15=k .4^{r}\) \(\Rightarrow 3 \times 5=k .4^{r}\)
\(\Rightarrow k .2^{r} \times 5=k .4^{r} \ldots . . .\)
\(\left(\because 3=k \cdot 2^{r}\right)\)
\(\Rightarrow 2^{r} \cdot 5=2^{2 r}\)
\(\Rightarrow 2^{r}=5\)
\(\Rightarrow r=\log _{2} 5\)

\(6\left(\log _{x} 2-\log _{4} x\right)+7=0\)
or \(\left(\log _{x} 2-\frac{1}{2} \log _{2} x\right)+7=0\)
\(6\left(\frac{1}{y}-\frac{y}{2}\right)+7=0\)
(where \(y=\log _{2} x\) )
or \(6\left(\frac{2-y^{2}}{2 y}\right)+7=0\)
or \(3\left(\frac{2-y^{2}}{y}\right)+7=0\)
\(6-3 y^{2}+7 y=0\)
\(3 y^{2}-7 y-6=0\)
or \(y^{2}+2 y-9 y-6=0\)
\((y-3)(3 y+2)=0\)
or \(y=3\) or \(y=-\frac{2}{3}\) \(\Rightarrow \log _{2} x=3\) or \(-\frac{2}{3}\)
or \(x=8\) or \(x=2^{-\frac{2}{3}}\)

\(\ln \left(\frac{a+b}{3}\right)=\frac{\ln a+\ln b}{2}\)
\(\ln \left(\frac{a+b}{3}\right)=\frac{\ln a b}{2}\)
\(\ln \left(\frac{a+b}{3}\right)=\ln \sqrt{a b}\)
Therefore \(\frac{a+b}{3}=\sqrt{a b}\)
\(a+b=3 \sqrt{a b}\)
\(\frac{a+b}{\sqrt{a b}}=3\)
\(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=3\)
Squaring both sides give us
\((\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}})^{2}=9\)
\(\frac{a}{b}+\frac{b}{a}+2=9\)
\(\frac{a}{b}+\frac{b}{a}=7\)

\(\log _{2} x+\log _{2} y \geq 6\)
\(\Rightarrow \log _{2} x y \geq 6,[\because \log a+\log b=\log (a b)]\)
\(\Rightarrow x y \geq 2^{6}=64\)....(i)
Also for \(\log x\) and \(\log y\) to be defined we must have, \(x>0 \&\) \(y>0\)
Now using, \(A . M . \geq G . M\)
\(\Rightarrow \frac{x+y}{2} \geq \sqrt{x y}\)
\(\Rightarrow \frac{x+y}{2} \geq \sqrt{64}=8 \quad \ldots\{\)
From 1\(\}\)
\(\Rightarrow x+y \geq 16\)

\(\log _{6} 9-\log _{9} 27+\log _{8} x=\log _{64} x-\log _{6} 4\)
\(\log _{8} x-\log _{64} x=\log _{9} 27-\log _{6} 4-\log _{6} 9\)
\(\log _{8} x-\log _{8^{2}} x=\log _{3^{2}} 27-\log _{6} 36\)
\(\log _{8} x-\frac{1}{2} \log _{8} x=\frac{1}{2} \log _{3} 27-2\)
\(\Rightarrow \frac{1}{2} \log _{8} x=\frac{3}{2}-2\)
\(\Rightarrow \log _{8} x=-1\)
\(\Rightarrow 8^{-1}=x\)
or \(x=\frac{1}{8}\)
Given eqn is defined for \(x>0\)
Hence, \(x=\frac{1}{8}\) is solution of given equation.

\(\log x+\log y=\log x-\log y+\log 4=0\)
\(\log y=-\log y+\log 4 \quad \ldots\) (Canceling \(\log\) x on both sides \()\)
\(2 \log y=\log 4\)
\(\log y^{2}=\log 2^{2}\)
\(y=2\)
\(\log 2 x=2\)
Assuming base 10
\(2 x=10^{2}\)
\(x=50\)

\(\left(\frac{1}{2}\right)^{x^{2}-2 x}<\frac{1}{4}\)
Taking logarithm on both sides, we get \(\left(x^{2}-2 x\right) \log \left(\frac{1}{2}\right)<\log \left(\frac{1}{4}\right)=\log \left(\frac{1}{2}\right)^{2}=2 \log \left(\frac{1}{2}\right)\)
\(\Rightarrow\left(x^{2}-2 x\right)>2,\) since \(\log (1 / 2)<0\)
\(\Rightarrow(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))>0\)
\(\Rightarrow x>1+\sqrt{3}\) or \(x<1-\sqrt{3}\)
Therefore, \(x \in (-\infty, 1-\sqrt{3}) \cup(1+\sqrt{3}, \infty)\)

\(\log _{3} x+\log _{3} y=2+\log _{3} 2=2 \log _{3} 3+\log _{3} 2,\left[\because \log _{3} 3=1\right]\)
As, \([\log a+\log b=\log (a b)]\) and \(\left[\because x \log a=\log a^{x}\right]\)
\(\Rightarrow \log _{3} x y=\log _{3} 3^{2}+\log _{3} 2\)
\(\Rightarrow \log _{3} x y=\log _{3} 18\)
\(\Rightarrow x y=18\)
\(\log _{3}(x+y)=2\)
\(\Rightarrow x+y=3^{2}=9\)
On solving equation ( 1 ) \(\&\) ( 2 ), we get \(x=3, y=6\)