f) x-1>0 => x>1
log_{\frac{1}{2}}(x-1)+log_{\frac{1}{2}}(x-1)=4
log_{\frac{1}{2}}(x-1)^2=4
(\frac{1}{2})^4=(x-1)^2
x^2-2x+1=\frac{1}{16}
x^2-2x+\frac{15}{16}=0
\Delta=(-2)^2-4*\frac{15}{16}=4-\frac{15}{4}=4-3\frac{3}{4}=\frac{1}{4}
\sqrt\Delta=\sqrt{\frac{1}{4}}=\frac{1}{2}
x_1=\frac{2-\frac{1}{2}}{2}=\frac{3}{2}*\frac{1}{2}=\frac{3}{4} nie spełnia warunków zadania
x_2=\frac{2+\frac{1}{2}}{2}=\frac{5}{2}*\frac{1}{2}=\frac{5}{4}=1\frac{1}{4}
g) x>0
log_3x+log_3\sqrt{x}=3
log_3x\sqrt{x}=3
3^3=x\sqrt{x}
x\sqrt{x}=3^3 |^2
x^2*x=(3^3)^2
x^3=(3^2)^3
x=3^2
x=9
h) x>0
log_5\sqrt{x}-log_5x=-1
log_5\frac{\sqrt{x}}{x}=-1
log_5x^{\frac{1}{2}-1}=-1
log_5x^{-\frac{1}{2}}=-1
log_5(\frac{1}{x})^{\frac{1}{2}}=-1
5^{-1}=\sqrt{\frac{1}{x}}
\frac{1}{5}=\sqrt{\frac{1}{x}} |^2
\frac{1}{25}=\frac{1}{x}
x=25
i) x>0
log_2(log_{\pi}x)=0
2^0=log_{\pi}x
log_{\pi}x=1
\pi^1=x
x=\pi