Question

look at the attachement and solve the asked questions fast please

Answer 1

74
$f(x)=\frac{1+x}{1-x}\\\\f(\frac{2x}{1+x^2})=Log[\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}]\\\\=Log[\frac{(1+x)^2}{(1-x)^2}]\\\\=2Log[\frac{1+x}{1-x}]=2f(x)$
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75.
$let\ k=(x-y)Log_a\ 3=(y-z)log_b\ 3=(z-x)log_c\ 3\\\\\frac{x-y}{k}+\frac{y-z}{k}+\frac{z-x}{k}=\frac{1}{Log_a\ 3}+\frac{1}{Log_b\ 3}+\frac{1}{Log_c\ 3}\\\\0=Log_3\ a+log_3\ b+log_3\ c\\\\0=Log_3\ abc\\\\abc=3^0=1\\\\=>5^{abc}=5$
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76.
$x = log_a\ bc=log_ab+log_ac;\\y=Log_b\ ca=Log_bc+Log_ba;\\z = log_c\ ab=log_ca+log_cb\\\\xy=(log_ab+log_ac)(Log_bc+Log_ba)\\.\ \ =log_ab\ Log_bc+1+Log_ac\ Log_bc+log_ba\ log_ac\\.\ \ =Log_ac+1+Log_ac\ Log_bc+Log_bc\\\\xyz=(1+Log_ac+Log_ac\ Log_bc+Log_bc)(log_ca+log_cb)\\.\ \ \ =log_ca+log_cb+1+Log_ab+Log_bc+Log_ac+Log_ba+1\\.\ \ \ =2+log_c\ ab+log_a\ bc+log_b\ ac\\.\ \ \ =2+x+y+z$

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77.
$log_4(x-1)=Log_2(x-3)=k\\\\x-1=4^k=2^{2k}=(2^k)^2\\x-3=2^k\\\\x-1=(x-3)^2\\=>x^2-7x+10=0\\=>as\ discriminant=7^2-4*10>0,there\ are2\ solutions$
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78.    x ≠ π/2 or π  as log cos x  and log sin 2x exist.

$let\ Log_{0.1}\ Sin2x=z\\Sin2x=0.1^z=10^{-z}\\z=-Log_{10}\ Sin2x\\\\Log_{0.1}\ Sin2x +log_{10}\ cosx=log_{10}\ \frac{1}{\sqrt3}\\\\-Log_{10}\ sin2x+log_{10}\ cosx=log_{10}\ \frac{1}{\sqrt3}\\\\Log_{10}\ cos\ x/Sin\ 2x=log_{10}\ \frac{1}{\sqrt3}\\\\\frac{1}{2sin\ x}=\frac{1}{\sqrt3}\\\\x=sin^{-1}\sqrt3/2=\pi/3$