Logarithmus mit Variable in der Potenz usw.

Question

9*5^2x=65,7

7+0,13^x=8,5

11,4^2x+3=0,9

Wie rechne ich diese Logarithmen aus?

Answer 1

9*5^(2x)=65,7

5^(2x) = 7,3

ln5^(2x)= ln7,3

2x*ln5 = ln7,3

x= (ln7,3/ln5)/2 = ...

Answer 2

Aloha :)

$$\left.9\cdot5^{2x}=65,7\quad\right|\;:9$$$$\left.5^{2x}=\frac{65,7}{9}=7,3\quad\right|\;\ln(\dots)$$$$\left.2x\ln5=\ln7,3\quad\right|\;:2\ln5$$$$\left.x=\frac{\ln7,3}{(2\ln5)}\approx0,6176\quad\right.$$

$$\left.7+0,13^x=8,5\quad\right|\;-7$$$$\left.0,13^x=1,5\quad\right|\;\ln(\dots)$$$$\left.x\ln0,13=\ln1,5\quad\right|\;:\ln0,13$$$$\left.x=\frac{\ln1,5}{\ln0,13}\approx-0,1987\quad\right.$$

$$\left.11,4^{2x+3}=0,9\quad\right|\;\ln(\dots)$$$$\left.(2x+3)\ln11,4=\ln0,9\quad\right|\;:\ln11,4$$$$\left.2x+3=\frac{\ln0,9}{\ln11,4}\quad\right|\;-3$$$$\left.2x=\frac{\ln0,9}{\ln11,4}-3\quad\right|\;:2$$$$\left.x=\frac{1}{2}\left(\frac{\ln0,9}{\ln11,4}-3\right)\approx-1,5216\quad\right.$$