We recently started with exponential functions, and I did this task for fun, but I apparently did everything wrong. I just don't get why it is wrong. I am aware of some logarithmic properties like $\log(\frac{x}{y})= \log(x)-\log(y)$ and $\log(x\cdot y)=\log(x)+\log(y)$ (though i don't really understand why that works and if $x$ and $y$ have to be variables or parameters. And I also don't know when to use these properties.)
The task:
Find the intersection point of $g(x)=3\cdot0.4^x$ and $f(x)=0.5\cdot1.5^x$
This is how i did it (and the solution is wrong):
$0.5\cdot1.5^x=3\cdot0.4^x$
$(0.5\cdot1.5^x)\cdot\frac{1}{3}\cdot\frac{1}{1.5^x}=(3\cdot0.4^x)\cdot\frac{1}{3}\cdot\frac{1}{1.5^x}$
$\frac{1}{6}=\frac{0.4^x}{1.5^x}$
$\frac{1}{6}=(\frac{0.4}{1.5})^x$
$\frac{1}{6}=x\cdot \log(\frac{4}{15})$
$\frac{1}{6\cdot \log(\frac{4}{15})}=x$
$-0,290$, which is terribly wrong.
The correct solution/approach:
$0.5\cdot1.5^x=3\cdot 0.4^x$
$\log(0.5\cdot 1.5^x)=\log(3\cdot 0.4^x)$
$\log(0.5)+\log(1.5^x)=\log(3)+\log(0.4^x)$
$\log(0.5)+x\cdot \log(1.5)=\log(3)+x\cdot \log(0.4)$
$\log(0.5)-\log(3)=x\cdot \log(0.4)-x\cdot \log(1.5)$
$\log(0.5)-\log(3)=x\cdot (\log(0.4)-\log(1.5))$
$\frac{\log(0.5)-\log(3)}{\log(0.4)-\log(1.5)}=x=1,36$
I would appreciate if someone could tell me why I wasn't able to divide and basically do the first approach.
The issue happens when you jump from $$\frac{1}{6} = \left (\frac{4}{15} \right )^{x}$$ to $$\frac{1}{6} = x\ln\left (\frac{4}{15} \right ).$$ In this step you have taken the logarithm of the right hand side and only the right hand side. What you need to do is take the logarithm of both sides $$\ln \left (\frac{1}{6} \right ) = x\ln\left (\frac{4}{15} \right ).$$ and then divide to obtain $$x = \frac{\ln(1/6)}{\ln(4/15)} \approx 1.36$$