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[koliber1000]

\(\displaystyle{ a) log _{4}(x+12)*log _{x}2= 1}\)
\(\displaystyle{ b) log(x-1)+ log(x-2)= log(x+2)}\)

[gajatko]

a)
\(\displaystyle{ \log _{4}(x+12)*\log _{x}2= 1\\
D=(0, 1) \cup (1, )}\)

Z tw. o zamianie podstaw
\(\displaystyle{ \log _4 (x+12)*\frac{\log_4 2}{\log _4 x}= 1 \\
{1\over 2} \log _4 (x+12) = \log _4 x\\
\sqrt{x+12}=x\\
x^2-x-12=0\\
x=4}\)
b)
\(\displaystyle{ D=(2,\infty)\\
\log(x-1)+ \log(x-2)= \log(x+2)\\
\log(x-1)(x-2)= \log(x+2)\\
x^2-2x-x+2=x+2\\
x=4}\)