2020/005) Compare $\log_23$ and $\log_35$

We have $2^3=8 < 9 = 3^2$
Hence $2^\frac{3}{2} < 3$
Or $\log_23 > \frac{3}{2}\cdots(1)$
$3^3=27 > 25 = 5^2$
Hence $3^\frac{3}{2} > 5$
Or $\log_35 <  \frac{3}{2}\cdots(2)$
Using (1) and (2) $\log_23 > \log_35$

2020/004) Prove that $\log_abc.\log_bca.\log_abc = 2 + \log_abc+\log_bca+\log_abc$

Solution
Let $\log_ab= x\cdots(1)$
$\log_bc=y\cdots(2)$
$\log_ca=z\cdots(3)$

So we have $xyz= \log_ab.\log_bc.\log_ca=1 \cdots(4)$
$\log_abc= \log_ab + \log_ac = x + \frac{1}{z}= x + xy= x(1+y)\cdots(5)$
Similarly
$\log_bca= y(1+z)\cdots(6)$
And
$\log_bca= z(1+x)\cdots(7)$


Now $LHS= \log_abc.\log_bca.\log_abc$
$=x(1+y) . y(1+z) . z (1+x)$ from (5),(6),(7)
$= xyz(1+y)(1+z)(1+x)$
$= (1+y)(1+z)(1+x)$ from (4) as xyz=1
$= 1 + y + z + x + xy+yz + zx + xyz$
$= 1 + y + z + x + xy+yz + zx + 1$ from (4) as xyz=1
$= 2 + y + yz + x + xy+ z + zx$ Rearrangement of terms
$=2 + y(1+z) + x(1+y) + z(1+x)$
$=2 +\log_bca + \log_abc + \log_cab$
$=2 + \log_abc + \log_bca + \log_cab$  Rearrangement of terms
$=RHS$


Hence Proved

2020/003) If 1 < x < 2, Then Prove that $\frac{1}{\sqrt{x+2\sqrt{x-1}}}+ \frac{1}{\sqrt{x-2\sqrt{x-1}}} = \frac {2}{2-x}$

Solution

Let $\sqrt{x+2\sqrt{x-1}} = \sqrt{m} + \sqrt{n}$
Hence  $\sqrt{x-2\sqrt{x-1}} = \sqrt{m} - \sqrt{n}$
Without loss of generality let us assume $m >=n$
Square both sides to get  $x+2\sqrt{x-1} = m + n + 2\sqrt{mn}$
Comparing rational and surds on both sides we get
$x=m +n\cdots(1)$
$x-1= mn\cdots(2)$
From (1) and (2)
$(m-n)^2 = (m+n)^2-4mn = x^2-4(x-1) = (2-x)^2$ we chose 2-x as $x < 2$
or $m-n=2-x\cdots(3)$
From (3) and (1) we get $m=1, n = x-1$
Hence we get
$\frac{1}{\sqrt{x+2\sqrt{x-1}}}+ \frac{1}{\sqrt{x-2\sqrt{x-1}}}$
$= \frac{1}{\sqrt{m}+\sqrt{n}}+ \frac{1}{\sqrt{m}-\sqrt{n}}$
$= \frac{\sqrt{m}+\sqrt{n}+ \sqrt{m}-\sqrt{n}} {m-n}$
$= \frac{2\sqrt{m}} {m-n}$
$= \frac{2\sqrt{1}} {2-x}$
$= \frac{2} {2-x}$

2020/002) a,b,c are in AP, x,y,z are in HP and ax,by,cz are in GP then prove that $\frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a}$

We have $x,y,z$ are in HP so
$\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$
Squaring both sides
$\frac{1}{x^2} + \frac{1}{z^2} + \frac{2}{xz} = \frac{4}{y^2}$
Or $\frac{1}{x^2} + \frac{1}{z^2} = \frac{4}{y^2} - \frac{2}{xz}$
Or $\frac{x^2+z^2}{x^2z^2} = \frac{4}{y^2} - \frac{2}{xz}$
Or $\frac{x^2+z^2}{xz} = \frac{4xz}{y^2} - 2$\
Or $\frac{x}{z} + \frac{z}{x} = \frac{4xz}{y^2} - 2\cdots(1)$
As ax,by,cz are in GP
$axcz = b^2y^2$
Or $\frac{xz}{y^2} = \frac{b^2}{ac}$
Putting above in (1)
$\frac{x}{z} + \frac{z}{x} = \frac{4b^2}{ac} - 2\cdots(2)$
As a,b,c are in AP we have $2b= a + c$
Putting in (2) we get
$\frac{x}{z} + \frac{z}{x} = \frac{(a+c)^2}{ac} - 2$
Or $\frac{x}{z} + \frac{z}{x} = \frac{(a+c)^2-2 ac} {ac}$
Or $\frac{x}{z} + \frac{z}{x} = \frac{a^2+c^2} {ac}$
Or $\frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a}$
proved

2020/001) a,b,c are in AP, b,c,d are in GP, and c,d,e are in HP, prove that a,c,e are in GP

Solution

We are given a,b,c are in AP so
$2b= a +c \cdots(1)$
b,c,d are in GP so
$bd=c^2\cdots(2)$
c,d,e are in HP, so
$\frac{1}{c} + \frac{1}{e} = \frac{2}{d}$
or $\frac{e+c}{ce} = \frac{2}{d}\cdots(3)$
or $d = \frac{2ce}{e+c}$
From(2)
$c^2 = bd$
or $2c^2 = 2bd = (a+c) \frac{2ce}{e+c}$ putting from (1) and (3)
or $c^2(e+c) = ce(a+c)$
or $c(e+c) = e(a+c)$
or $c^2= ae$

Hence  a,c,e are in GP