3^2x = 4*5 ^(x+3) lösen

Question

NO  Aufgabe: 3^2x = 4*5^(x+3)

Problem/Ansatz:

… Ich komme nicht auf ein Ergebnis

Answer 1

\(\begin{aligned} 3^{2x} & =4\cdot5^{x+3} &  & |\log\\ 2x\cdot\log3 & =\log4+\left(x+3\right)\log5\\ 2x\cdot\log3 & =\log4+x\log5+3\log5 &  & |-x\log5\\ 2x\cdot\log3-x\log5 & =\log4+3\log5\\ x\left(2\log3-\log5\right) & =\log4+3\log5 &  & |:\left(2\log3-\log5\right)\\ x & =\frac{\log4+3\log5}{2\log3-\log5}\\ & =\frac{\log\left(4\cdot5^{3}\right)}{\log\left(\frac{3^{2}}{5}\right)}\\ & =\frac{\log500}{\log\frac{9}{5}} \end{aligned}\)

Answer 2

3^(2·x) = 4·5^(x + 3)

9^x = 4·125·5^x

9^x = 500·5^x

9^x/5^x = 500

(9/5)^x = 500

x = LN(500)/LN(9/5) = 10.57

Answer 3

Aloha :)

$$\left.3^{2x}=4\cdot5^{x+3}\quad\right|\quad\ln(\cdots)$$$$\left.\ln(3^{2x})=\ln(4\cdot5^{x+3})\quad\right|\quad\ln(a\cdot b)=\ln(a)+\ln(b)$$$$\left.\ln(3^{2x})=\ln(4)+\ln(5^{x+3})\quad\right|\quad\ln(a^b)=b\cdot\ln(a)$$$$\left.2x\cdot\ln(3)=\ln(4)+(x+3)\ln(5)\quad\right|\quad\text{rechts die Klammer ausrechnen}$$$$\left.x\cdot2\ln(3)=\ln(4)+x\ln(5)+3\ln(5)\quad\right|\quad-x\ln(5)$$$$\left.x\cdot2\ln(3)-x\ln(5)=\ln(4)+3\ln(5)\quad\right|\quad\text{links \(x\) ausklammern}$$$$\left.x\cdot(2\ln(3)-\ln(5))=\ln(4)+3\ln(5)\quad\right|\quad:\,(2\ln(3)-\ln(5))$$$$x=\frac{\ln(4)+3\ln(5)}{2\ln(3)-\ln(5)}\approx10,5729$$