Exercise 2.12 - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 2.12
Question 1.
Let $\mathrm{b}>0$ and $\mathrm{b} \neq 1$. Express $\mathrm{y}=\mathrm{b}^{\mathrm{x}}$ in logarithmic form. Also state the domain and range of the logarithmic function.
Solution:
Given $y=\mathrm{b}^{\mathrm{x}} \Rightarrow \log _{\mathrm{b}} \mathrm{y}=\mathrm{x}, \mathrm{x} \in \mathrm{R}$ with range $(0, \infty)(-\infty, \infty)$
Question 2.
Compute $\log _9 27-\log _{27} 9$.
Solution:
Let $\log _9 27=\mathrm{x} \Rightarrow 27=9^{\mathrm{x}} \Rightarrow 3^3=\left(3^2\right)^{\mathrm{x}}=3^{2 \mathrm{x}}$
$\Rightarrow 2 \mathrm{x}=3 \Rightarrow \mathrm{x}=3 / 2$
Let $\log _{27} 9=x$
$
\begin{aligned}
9 & =27^x \\
3^2 & =\left(3^3\right)^x \Rightarrow 3^2=3^{3 x} \\
3 x & =2 \Rightarrow x=2 / 3 \\
\therefore \log _9 27-\log _{27} 9 & =\frac{3}{2}-\frac{2}{3}=\frac{9-4}{6}=\frac{5}{6}
\end{aligned}
$
Question 3.
Solve $\log _8 \mathrm{x}+\log _4 \mathrm{x}+\log _2 \mathrm{x}=11$
Solution:

$
\begin{aligned}
& \log _8 x=\frac{1}{\log _x 8}=\frac{1}{\log _x 2^3}=\frac{1}{3 \log _x 2} \\
& \log _4 x=\frac{1}{\log _x 4}=\frac{1}{\log _x 2^2}=\frac{1}{2 \log _x 2} \\
& \log _2 x=\frac{1}{\log _x 2} \\
& \therefore \log _8 x+\log _4 x+\log _2 x=11 \\
& \Rightarrow \frac{1}{3 \log _x 2}+\frac{1}{2 \log _x 2}+\frac{1}{\log _x 2}=11 \\
& \Rightarrow \quad \frac{1}{\log _x 2}\left[\frac{1}{3}+\frac{1}{2}+1\right]=11 \\
& \Rightarrow \quad \frac{1}{\log _x 2}\left[\frac{2+3+6}{6}\right]=11 \\
& \Rightarrow \quad \frac{1}{\log _x 2}=11 \times \frac{6}{11}=6 \\
& \text { (i.e.,) }=6 \\
& x=2^6=64 \\
&
\end{aligned}
$
Question 4.
Solve $\log _4 2^{8 x}=2^{\log _2 8}$
Solution:
$
\begin{aligned}
& \text { Let } \log _4 2^{8 x}=\mathrm{A} \Rightarrow 2^{8 x}=4^A \\
& 2^{8 x}=\left(2^2\right)^{\mathrm{A}}=2^{2 \mathrm{~A}} \\
& \Rightarrow 2 \mathrm{~A}=8 x \Rightarrow \mathrm{A}=\frac{8 x}{2}=4 x \\
& \text { LHS }=4 x \\
& \text { RHS }=2^{\log _2 8}=2^3=8 \\
& {\left[\log _2 8=\log _2 2^3=3 \log _2 2=3 \times 1=3\right]} \\
& \therefore 4 x=8 \\
& x=8 / 4=2 \\
&
\end{aligned}
$

Question 5.
If $a^2+b^2=7 a b$ show that $\log \frac{a+b}{3}=\frac{1}{2}(\log a+\log b)$
Solution:
$
\begin{aligned}
& a^2+b^2=7 a b \\
\therefore \quad & a^2+b^2+2 a b=7 a b+2 a b=9 a b \\
& (\text { i.e., })(a+b)^2=9 a b \\
\therefore \quad & a+b=\sqrt{9 a b}=3 \sqrt{a b} \\
\Rightarrow & \frac{a+b}{3}=\sqrt{a b}=(a b)^{\frac{1}{2}}
\end{aligned}
$
Taking log on both sides we get
$
\log \frac{a+b}{3}=\log (a b)^{1 / 2}=\frac{1}{2} \log a b=\frac{1}{2}[\log a+\log b]
$
Question 6.
Prove that $\log \frac{a^2}{b c}+\log \frac{b^2}{c a}+\log \frac{c^2}{a b}=0$
Solution:
$
\begin{aligned}
& {[\log a+\log b+\log c=\log a b c] } \\
& \text { LHS }=\log \frac{a^2}{b c}+\log \frac{b^2}{c a}+\log \frac{c^2}{a b}=\log \frac{a^2}{b c} \times \frac{b^2}{c a} \times \frac{c^2}{a b} \\
&=\log \frac{a^2 b^2 c^2}{a^2 b^2 c^2}=\log 1=0=\text { RHS }
\end{aligned}
$

Another methods
$
\begin{aligned}
& \log \frac{a^2}{b c}=\log a^2-\log b c=2 \log a-\log b-\log c \\
& \log \frac{b^2}{c a}=\log b^2-\log c a=2 \log b-\log c-\log a \\
& \log \frac{c^2}{a b}=\log c^2-\log a b=2 \log c-\log a-\log b \\
& \text { (1) }+(2)+(3)=2 \log a-\log b-\log c+2 \log b-\log c-\log a \\
& +2 \log c-\log a-\log b=0=\text { RHS } \\
&
\end{aligned}
$
Question 7.
Prove that $\log 2+16 \log \frac{16}{15}+12 \log \frac{25}{24}+7 \log \frac{81}{80}=1$

Solution:
$
\begin{aligned}
& \text { LHS }=\log 2+16[\log 16-\log 15]+12[\log 25-\log 24]+7[\log 81-\log 80] \\
& \left\{\begin{array}{r}
\text { Now } 16=2^4 ; 15=3 \times 5 \\
25=5^2 ; 24=2^3 \times 3 \\
81=3^4 ; 80=2^4 \times 5
\end{array}\right\} \\
= & \log 2+16\left[\log 2^4-\log 3 \times 5\right]+12\left[\log 5^2-\log 2^3 \times 3\right]+7\left[\log 3^4-\log 2^4 \times 5\right] \\
= & \log 2+16[4 \log 2-\log 3-\log 5]+12[2 \log 5-3 \log 2-\log 3] \\
& +7[4 \log 3-4 \log 2-\log 5] \\
= & \log 2+64 \log 2-16 \log 3-16 \log 5+24 \log 5-36 \log 2-12 \log 3+28 \log 3 \\
& -28 \log 2-7 \log 5 \\
= & \log 2[1+64-36-28]+\log 3[-16-12+28]+\log 5[-16+24-7] \\
= & \log 2(1)+\log 3(0)+\log 5(1) \\
= & \log 2+\log 5=\log 2 \times 5=\log 10=1=\text { RHS }
\end{aligned}
$
Question 8.
Prove that $\log _{a^2} a \log _{b^2} b \log _{c^2} c=\frac{1}{8}$
Solution:
Let $\log _{a^2} a=x$
$
\begin{array}{cc}
\Rightarrow & a=\left(a^2\right)^x=a^{2 x} \\
\Rightarrow \quad & a^{2 x}=a^1 \Rightarrow 2 x=1 \Rightarrow x=1 / 2 \\
& \log _{b^2} b=1 / 2 \text { and } \log _{c^2} c=1 / 2 \\
& \text { LHS }=\log _{a^2} a \log _{b^2} b \log _{c^2} c=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}=\text { RHS }
\end{array}
$
Question 9.
Prove that $\log a+\log a^2+\log a^3+\ldots+\log a^n=\frac{n(n+1)}{2} \log a$
Solution:

$
\begin{aligned}
& \text { LHS }=\log a+\log a^2+\log a^3+\ldots+\log a^n \\
& =\log a+2 \log a+3 \log a+\ldots .+n \log a \\
& =\log a(1+2+3+\ldots n) \\
& =\log a\left[\frac{n(n+1)}{2}\right]=\frac{n(n+1)}{2} \log a=\text { RHS } \\
&
\end{aligned}
$
Question 10
If $\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}$ then prove that $x y z=1$
Solution:
Let $\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}=k$ (say)
Now $\frac{\log x}{y-z}=k \Rightarrow \log x=k(y-z) \Rightarrow x=e^{k(y-z)}$
$
\begin{aligned}
& \frac{\log y}{z-x}=k \Rightarrow \log y=k(z-x) \Rightarrow y=e^{k(z-x)} \\
& \frac{\log z}{x-y}=k \Rightarrow \log z=k(x-y) \Rightarrow z=e^{n(x-y)}
\end{aligned}
$
Now LHS $=x y z=e^{k(y-z)} e^{k(z-x)} e^{k(x-y)}$
$
\begin{aligned}
& =e^{k(y-z+z-x+x-y)} \\
& =e^{k(0)}=e^0=1=\text { RHS }
\end{aligned}
$
Question 11
Solve $\log _2 x-3 \log _{1 / 2} x=6$

Solution:
$
\begin{aligned}
& \log _2 x=\log _{1 / 2} x \cdot \log _2 1 / 2 \\
& =\log _{1 / 2} x \log _2 1-2 \\
& =\log _{1 / 2} x(-1)=-\log _{1 / 2} x \\
& \therefore \log _2 x-3 \log _{1 / 2} x=6 \Rightarrow-\log _{1 / 2} x-3 \log _{1 / 2} x=6 \\
& \text { (i.e.,) } \quad-\log _{1 / 2} x-\log _{1 / 2} x^3=0 \\
& \Rightarrow \quad-\log _{1 / 2}\left(x \times x^3\right)=6 \\
& \Rightarrow \quad \log _{1 / 2} x^4=-6 \\
& \Rightarrow \quad x^4=\left(\frac{1}{2}\right)^{-6}=\frac{1}{2^{-6}}=2^6=64 \\
& \Rightarrow \quad x=(64)^{1 / 4}=\left[(64)^{1 / 2}\right]^{1 / 2}=(8)^{1 / 2}=2 \sqrt{2} \\
&
\end{aligned}
$
Question 12 .
Solve $\log _{5-x}\left(x^2-6 x+65\right)=2$
Solution:
$
\begin{aligned}
& \log _{5-x}\left(x^2-6 x+65\right)=2 \\
& \Rightarrow x^2-6 x+65=(5-x)^2 \\
& x^2-6 x+65=25+x^2-10 x \\
& \Rightarrow x^2-6 x+65-25-x^2+10 x=0 \\
& 4 x+40=0 \Rightarrow 4 x=-40 \\
& x=-10
\end{aligned}
$