Malliratkaisuja

456.
\ln\left(5x-1\right)=\ln\left(2x\right)
\text{Määrittelyjoukko: }\ 5x-1>0\ \text{ja}\ 2x>0
x>\frac{1}{5}
5x-1=2x
3x=1
x=\frac{1}{3}
 
b)
2\log_2x=\log_2\left(3x+4\right)
Määrittelyjoukko:
x>0\ \text{ja}\ 3x+4>0\ \text{eli}\ x>-\frac{4}{3}
x>0
2\log_2x=\log_2\left(3x+4\right)
\log_2x^2=\log_2\left(3x+4\right)
x^2=3x+4
\left(x=-1\right)\ \text{tai}\ x=4
v: x = 4
 
 
 
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D ln x = 1/x todistus:

f\left(x\right)=\ln x
Derivoidaan kohdassa x = x_0
f'\left(x_0\right)=\lim_{x\rightarrow}x_0\left(\frac{\ln x-\ln x_0}{x-x_0}\right)
y=\ln x{,}\ \ \ y_0=\ln x_0{,}\ \ \text{joten}\ \ x=e^y\ \text{ja}\ x_0=e^{y_0}
 
f'\left(x\right)=\lim_{x\rightarrow x_0}\left(\frac{\ln x-\ln x_0}{x-x_0}\right)=\lim_{x\rightarrow x_0}\left(\frac{y-y_0}{e^y-e^{y_0}}\right)=\lim_{x\rightarrow x_0}\left(\frac{1}{\frac{e^y-e^{y_0}}{y-y_0}}\right)  || \lim_{y\rightarrow y_0}\left(\frac{e^y-e^{y_0}}{y-y_0}\right)=e^{y_0}
=\lim_{x\to x_0}\left(\frac{1}{e^{y_0}}\right)=\lim_{x\to x_0}\left(\frac{1}{x_0}\right)=\frac{1}{x_0}