I dont remember how to do this problem and being summer, I dont have my textbook to help.
Solve: 4^(x+2)=9^(x-1)
Hello !
Just use logarithms.
Remember this property : $\displaystyle \ln(a^b)=b \ln(a)$
$\displaystyle \ln(4^{x+2})=\ln(9^{x-1})$
$\displaystyle (x+2) \ln(4)=(x-1) \ln(9)$
Can you do the rest ? (you can simplify again in the end by writing $\displaystyle 4=2^2 \quad \& \quad 9=3^2$)
Whatever base you choose, the rules are the same
$\displaystyle \ln$ denotes the natural logarithm and is like an *universal* base :
$\displaystyle \ln(e)=1$
Continuing from $\displaystyle (x+2) \ln(4)=(x-1) \ln(9)$ :
$\displaystyle x \ln(4)+2 \ln(4)=x \ln(9)-\ln(9)$
Group the x together :
$\displaystyle x \ln(9)-x \ln(4)=2 \ln(4)+\ln(9)$
$\displaystyle x(\ln(9)-x \ln(4))=2 \ln(4)+\ln(9)$
$\displaystyle \implies x=\frac{2 \ln(4)+\ln(9)}{\ln(9)-\ln(4)} \quad (1)$
Now, if you take the logarithm in base 2, you'll have $\displaystyle \log_2(4)=2$ because $\displaystyle 4=2^2$. Actually, it's $\displaystyle \log_2(4)=\log_2(2^2)=2 \underbrace{\log_2(2)}_{=1}=2$.
But in a common logarithm, you have :
$\displaystyle \ln(4)=\ln(2^2)=2 \ln(2)$, same goes for $\displaystyle \ln(9)=\ln(3^2)=2 \ln(3)$
Substituting in (1) :
$\displaystyle x=\frac{4 \ln(2)+2 \ln(3)}{2 \ln(3)-2 \ln(2)}$
Simplify by 2 :
$\displaystyle \boxed{x=\frac{2 \ln(2)+\ln(3)}{\ln(3)-\ln(2)}}$
You gotta learn how to use the natural logarithm. Using bases just sometimes avoid difficulties things easier, but in exercises like this one, it brings difficulties !