Exercise 2.12-Additional Questions - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Solve $\log _{16} x+\log _4 x+\log _2 x=7$

Solution:
$
\begin{aligned}
\log _{16} x & =\frac{1}{\log _x 16}=\frac{1}{\log _x 2^4}=\frac{1}{4 \log _x 2} \\
\log _4 x & =\frac{1}{\log _x 4}=\frac{1}{\log _x 2^2}=\frac{1}{2 \log _x 2} \\
& \Rightarrow-\log _{16} x+\log _4 x+\log _2 x=7 \\
\Rightarrow \quad & \frac{1}{4 \log _x 2}+\frac{1}{2 \log _x 2}+\frac{1}{\log _x 2}=7 \\
\Rightarrow \quad & \frac{1}{\log _x 2}\left[\frac{1}{4}+\frac{1}{2}+1\right]=7 \\
\text { (i.e.,) } & \frac{1}{\log _x 2}\left[\frac{1+2+4}{4}\right]=7 \\
\text { (i.e.,) } & \frac{1}{\log _x 2}\left[\frac{7}{4}\right]=7 \Rightarrow \frac{1}{\log _x 2}=7 \times \frac{4}{7}=4
\end{aligned}
$

Question 2.
Prove that $\log _5 1125=2 \log _5 6-\frac{1}{2} \log _5 16+6 \log _{49} 7$
Solution:
$
\begin{aligned}
\text { R.H.S. } & =2 \log _5 6-\frac{1}{2} \log _5 16+6 \log _{49} 7 \\
& =\log _5 6^2-\log _5 16^{\frac{1}{2}}+\log _{49} 7^6 \\
& =\log _5 36-\log _5 4+\log _{49}\left(7^2\right)^3 \\
& =\log _5 \frac{36}{4}+3 \log _{49} 7^2 \\
& =\log _5 9+3 \log _{49} 49=\log _5 9+3 \times 1 \\
& =\log _5(9 \times 125)=\log _5 9+\log _5 125 \\
& =\log (9 \times 125)=\log _5 1125=\text { LHS }
\end{aligned}
$

Question 3.
Solve $\log _5 \sqrt{7 x-4}-\frac{1}{2}=\log _5 \sqrt{x+2}$

Solution:

Question 4.
If $a^2+b^2=23 a b$ prove that $\log \frac{a+b}{5}=\frac{1}{2}(\log a+\log b)$
Solution:
$
a^2+b^2=23 a b
$
Adding $2 a b$ on both sides
$
\begin{aligned}
& a^2+b^2+2 a b=23 a b+2 a b=25 a b \\
&(\text { i.e..) })(a+b)^2=25 a b \\
& \Rightarrow \quad a+b=\sqrt{25 a b}=5(a b)^{1 / 2} \\
& \therefore \frac{a+b}{5}=(a b)^{1 / 2}
\end{aligned}
$
Taking log on both sides we get
$
\log \frac{a+b}{5}=\log (a b)^{1 / 2}=\frac{1}{2}(\log a b)=\frac{1}{2}[\log a+\log b]
$