## myininaya 4 years ago for supercrazy: how to solve $3^{4x-7}=4^{2x+3}$

1. myininaya

Do you remember $\ln(x^r)=r \ln(x)$

2. myininaya

i got stood up

3. anonymous

lol

4. amistre64

those loger-epo rules make more sense when defined by calculus

5. anonymous

@Myini-Did I do something? o.o Sorry I will delete my comment.

6. myininaya

No this person wanted to know how to do this problem in chat and I said was going to post it here and she never came. :(

7. anonymous

@Myini-You did your work nevertheless, shows you're a sincere person :)

8. anonymous

Just to practice latex \begin{align}3^{4x-7}=4^{2x+3} &<=> (4x-7)ln(3) &&=(2x+3)ln(4) \\ &<=> x4ln(3)-7ln(3) &&= x2ln(4)+3ln(4) \\ &<=> x(4ln(3)-2ln(4)) &&= 3ln(4)+7ln(3) \\ & <=> x &&={ 3ln(4)+7ln(3) \over 4ln(3)-2ln(4)} \\ or\ & x={ln(4^3.3^7) \over ln(3^4.4^2) }&& = {ln(139968)\over ln(1296)}\approx 1.65\end{align}

9. myininaya

Pretty mkone!

10. myininaya

well i think i see something just a little off on your solution

11. myininaya

$4 \ln(3)-2 \ln(4)=\ln(3^4)-\ln(4^2)=\ln(\frac{3^4}{4^2})$

12. anonymous

I'm not getting what's off is it the presentation. If so I agree $$ln({3^4 \over 4^2})$$ looks better but then what about $$2ln({3^2 \over 4})$$ ?