How would I go about a problem like this?:
2^x = 3^(x-1)
I know that usually you try to get the same base but I don't think that's possible here.

Question

How would I go about a problem like this?:
2^x = 3^(x-1)
I know that usually you try to get the same base but I don't think that's possible here.

epoweritheta · Answer

bro why your question shows high ??? in red color ??

cocoacat · Answer

I don't know how to turn it off!

epoweritheta · Answer

oh...is it taking away you owl-money ??/

okay bring down the $$3^{x}$$ over to this side to the denominator

epoweritheta · Answer

try and tell me

cocoacat · Answer

I don't have owl money 0-0
(2^x)/(3^(x-1)) = 1

epoweritheta · Answer

now think how to solve ...

epoweritheta · Answer

bro keep thinking i will be back in 10 mins ...  have to eat :P

cocoacat · Answer

Ok!

unklerhaukus · Answer

$$2^x = 3^{x-1}$$Take the log to base 2 on both sides
$$\log_2(2^x) = \log_2(3^{x-1})$$Simplify the left hand side
$$x = \log_2(3^{x-1})$$
Apply the change of base formula $\boxed{\log_b(a)=\frac{\log_c(a)}{\log_c(b)}}$, (with $c = 3$ as the new base)
$$x =\frac{\log_3(3^{x-1})}{\log_3(2)}$$
simplify
$$\qquad \vdots$$

mathmath333 · Answer

yup

epoweritheta · Answer

ooohh.... god... 

bro  multiply and divide by 3 

yielding $$3*(2/3)^x=1$$ 

now you can solve it